Table of Contents

Quadratic Functions in Algebra (IB Mathematics Analysis and Approaches Standard Level)

A quadratic function is a polynomial function of degree 2. The general form of a quadratic function is:

f (x) = a x^{2} + b x + c

where:
  • 𝑎, 𝑏, and 𝑐 are constants,
  • 𝑎≠0,
  • 𝑥 is the variable.

Key Features of Quadratic Functions

1. Parabola:
The graph of a quadratic function is called a parabola.

It opens upwards if 𝑎>0 and downwards if 𝑎<0.

2. Vertex:
The vertex of the parabola is the highest or lowest point, depending on whether it opens downwards or upwards, respectively.

The vertex form of a quadratic function is:

f (x) = a (x - h)^{2} + k

where (ℎ,𝑘)is the vertex of the parabola.

The vertex can also be found using the formula:

h = - \frac{b}{2 a}

Substituting ℎ into the function to find 𝑘:

k = f (- \frac{b}{2 a})

3. Axis of Symmetry:

The axis of symmetry is the vertical line that passes through the vertex.

It has the equation:
x=ℎ
where ℎ is the x-coordinate of the vertex.

4. Roots (or Zeros):

The roots of the quadratic function are the values of 𝑥 that make 𝑓(𝑥)=0

They can be found using the quadratic formula:

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

The expression under the square root, b² - 4ac, is called the discriminant.

5. Discriminant:

The discriminant determines the nature of the roots.

if b² - 4ac > 0 , there are two distinct real roots.

if b² - 4ac = 0, there is one real root (a repeated root).

if b² - 4ac < 0, there are no real roots (the roots are complex)

Transformations of Quadratic Functions

Transformations can be used to modify the graph of a quadratic function. Common transformations include:

1. Vertical Translation:

Adding or subtracting a constant 𝑘 shifts the graph up or down.

f (x) = a x^{2} + b x + (c + k)

2. Horizontal Translation:

Adding or subtracting a constant ℎ to 𝑥 shifts the graph left or right.

f (x) = a (x - h)^{2} + k

3. Reflection:

Reflecting across the x-axis changes the sign of 𝑎.

f (x) = - a x^{2} - b x - c

4. Vertical Stretch or Compression:

Multiplying by a constant 𝑎 changes the width of the parabola.

if |a| > 1, the graph is stretched.

if 0 < |a| < 1, the graph is compressed.

Solving Quadratic Equations

Quadratic equations can be solved using several methods:

1. Factoring:

Write the quadratic equation in factored form:

a x^{2} + b x + c = (d x + e) (f x + g) = 0

Set each factor to zero and solve for 𝑥.

2. Completing the Square:

Rewrite the quadratic equation in vertex form by completing the square. Solve for 𝑥.

3. Quadratic Formula:

Use the quadratic formula:

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

Applications of Quadratic Functions

Quadratic functions have numerous applications, including:

1. Projectile Motion:

The path of a projectile under gravity is modeled by a quadratic function.

Example: y=ax² + bx +c, represents the height and 𝑥 the horizontal distance.

2. Optimization Problems:

Quadratic functions are used in optimization to find maximum or minimum values.

Example: Maximizing the area of a rectangle given a fixed perimeter.

3. Economics:

Quadratic functions model various economic phenomena such as profit, cost, and revenue functions.

Problem 1:

Consider a quadratic function given by the equation $f (x) = a x^{2} + b x + c$ . You are told that the vertex of the parabola represented by this equation occurs at (-3,4) and it passes through the point (2,-1). Determine the values of a,b, and c.

Solution:

Use vertex form of a quadratic equation.
A quadratic equation in vertex form is given by:
$f (x) = a (x - h)^{2} + k$

where (h,k) is the vertex of the parabola. Plugging in the vertex (-3,4) for (h,k), we get:

\begin{aligned} f (x) = a (x - (- 3))^{2} + 4 \\ y = a (x + 3)^{2} + 4 \end{aligned}

Substitute the point (2,-1) for (x,y) into the vertex form:
We know that f(2) = -1. Substitute 2 for x and -1 for y in the modified equation:

\begin{aligned} - 1 = a (2 + 3)^{2} + 4 \\ - 1 = a (5)^{2} + 4 \\ - 1 = 25 a + 4 \end{aligned}

Solve for 𝑎

\begin{aligned} 25 a + 4 = - 1 \\ 25 a = - 5 \\ 25a = \frac{5}{25} \\ 25a = - \frac{1}{5} \end{aligned}

Substitute 𝑎 back into the vertex form to find 𝑏 and 𝑐

Now that we know $a = - \frac{1}{5}$ , substitute 𝑎 back into the vertex form and expand it to standard form:

f (x) = - \frac{1}{5} (x + 3)^{2} + 4

expand and simplify $(x + 3)^{2}$

\begin{aligned} f (x) = - \frac{1}{5} (x + 3) (x + 3) + 4 \\ f (x) = - \frac{1}{5} (x^{2} + 6 x + 9) + 4 \\ f (x) = - \frac{1}{5} x^{2} - \frac{6}{5} x - \frac{9}{5} + 4 \\ f (x) = - \frac{1}{5} x^{2} - \frac{6}{5} x + \frac{11}{5} \end{aligned}

Conclusion:
Therefore, the values of 𝑎, 𝑏, and 𝑐 in the standard form $a x^{2} + b x + c$ of the given quadratic function are:

Here's the graph of the quadratic function $f (x) = a x^{2} + b x + c$ . It shows the vertex at (-3,4) and, also includes the point (2,-1) through which the parabola passes. The graph helps visualize the shape and positioning of the parabola relative to these key points.

Problem 2:

Consider a quadratic function defined by the equation $f (x) = a x^{2} + b x + c$ . The vertex of the parabola described by this equation is located at (1,-5) and it intersects the y-axis at (0,3). Determine the values of a,b, and c.

Solution:

Use vertex form of a quadratic equation.
The vertex form of a quadratic function is:
$f (x) = a (x - h)^{2} + k$

where (h,k) is the vertex. Substitute the vertex (1,-5) for (h,k), the equation becomes:

f (x) = a (x - 1)^{2} - 5

Use the y-intercept to find a

Since the parabola intersects the y-axis at (0,3). we know f (0)=3. Substitute 0 for 𝑥 and 3 for y in the equation:

\begin{aligned} 3 = a (0 - 1)^{2} - 5 \\ 3 = a (1) - 5 \\ 3 + 5 = a \\ a = 8 \end{aligned}

Substitute 𝑎 back into the vertex form

Now that we have a = 8, so substitute it back:

$f (x) = 8 (x - 1)^{2} - 5$

Expanding this to standard form:

\begin{aligned} f (x) = 8 (x^{2} - 2 x + 1) - 5 \\ f (x) = 8 x^{2} - 16 x + 8 - 5 \\ f (x) = 8 x^{2} - 16 x + 3 \end{aligned}

Conclusion:

Thus, the coefficients 𝑎a, 𝑏b, and 𝑐c in the standard form $a x^{2} + b x + c$ of the given quadratic function are:

$a = 8, b = - 16, c = 3$

Let’s Plot the graph for this Quadratic function..

Here's the graph of the quadratic function $f (x) = 8 x^{2} - 16 x + 3$ .The graph visually represents the vertex at (1,-5)
and the y-intercept at (0,3). This visualization helps in understanding the shape and key features of the parabola based on the problem's conditions.

Problem 3:

You are given a quadratic function described by the equation
$f (x) = a x^{2} + b x + c$ . This parabola has its vertex at (2,-7) and it crosses the x-axis at two points, one of which is (5,0). Determine the values of 𝑎, 𝑏, and 𝑐.

Solution:

Use the vertex form of a quadratic equation.
The vertex form of a quadratic function is:
$f (x) = a (x - h)^{2} + k$

where (h,k) is the vertex. Given the vertex (2,-7), the equation becomes:
$f (x) = a (x - 2)^{2} - 7$

Use the x-Intercept to find a
Since the parabola touches the x-axis at (5,0), we know f (5) = 0. Substitute 5 for x in the equation:

$\begin{aligned} 0 = a (5 - 2)^{2} - 7 \\ 0 = 9 a - 7 \\ 9 a = 7 \\ a = \frac{7}{9} \end{aligned}$

Substitute 𝑎 back into the vertex form
Now we have $a = \frac{7}{9}$ , so substitute it back:

$f (x) = \frac{7}{9} (x - 2)^{2} - 7$

Expanding this to standard form:

$\begin{aligned} f (x) = \frac{7}{9} (x^{2} - 4 x + 4) - 7 \\ f (x) = \frac{7}{9} x^{2} - \frac{28}{9} x + \frac{28}{9} - 7 \\ f (x) = \frac{7}{9} x^{2} - \frac{28}{9} x - \frac{35}{9} \end{aligned}$

Conclusion:
Thus, the coefficients 𝑎, 𝑏, and 𝑐 in the standard form $a x^{2} + b x + c$ of the given quadratic function are:

$a = \frac{7}{9}, b = - \frac{28}{9}, c = - \frac{35}{9}$

Let's plot the graph for this quadratic function to visualize the vertex and the x-intercept.

Here's the graph of the quadratic function
$f (x) = \frac{7}{9} x^{2} - \frac{28}{9} x - \frac{35}{9}$ .
It visually illustrates the vertex at (2,-7) and the x-intercept at (5,0).
This graph helps in understanding the properties of the parabola, especially how it interacts with the x-axis and its vertex characteristics.

Problem 4:
You are given a quadratic function described by the equation $f (x) = a x^{2} + b x + c$ .
The vertex of the parabola is located at (-2,3), and it crosses the x-axis at points (-5,0) and (1,0). Determine the values of a, b, and c.

Solution:

Use the vertex form of a quadratic equation.
The vertex form of a quadratic function is:
$f (x) = a (x - h)^{2} + k$

where (h,k) is the vertex. Given the vertex (-2,3), the equation becomes:
$f (x) = a (x + 2)^{2} + 3$

Use the x-axis intercepts.

Given the x-axis intercepts at (-5,0) and (1,0), we know f (-5) = 0 and f (1) = 0.
Substitute -5 for x and 0 for y into the equation:

For x = -5:

$\begin{aligned} 0 = a (- 5 + 2)^{2} + 3 \\ 0 = a (- 3)^{2} + 3 \\ 0 = 9 a + 3 \\ 9 a = - 3 \\ a = - \frac{1}{3} \end{aligned}$

For x = 1, let’s verify:

$\begin{matrix} 0 = - \frac{1}{3} (1 + 2)^{2} + 3 \\ 0 = - \frac{1}{3} (1 + 2)^{2} + 3 \\ 0 = - 3 + 3 \end{matrix}$

$0 = 0 (This confirms that: a = - \frac{1}{3} is correct.)$

Convert to standard form

Now we have $a = \frac{- 1}{2}$ , substitute back into the vertex form and expand:

$\begin{aligned} f (x) = - \frac{1}{3} (x + 2)^{2} + 3 \\ f (x) = - \frac{1}{3} (x^{2} + 4 x + 4) + 3 \\ f (x) = - \frac{1}{3} x^{2} - \frac{4}{3} x - \frac{4}{3} + 3 \\ f (x) = - \frac{1}{3} x^{2} - \frac{4}{3} x + \frac{5}{3} \end{aligned}$

Conclusion:

Thus, the coefficients a,b, and c in the standard form $a x^{2} + b x + c$ of the given quadratic function are:

$a = - \frac{1}{3}, b = - \frac{4}{3}, c = \frac{5}{3}$

Let's visualize this with a graph to see the vertex and the x-axis intercepts.

Here's the graph of the quadratic function $f (x) = - \frac{1}{3} x^{2} - \frac{4}{3} x + \frac{5}{3}$ . It displays the vertex at (-2,3) and the x-axis intercepts at (-5,0) and (1,0). This visualization helps understand how the parabola behaves with its vertex and intersection points clearly marked.

More Free Blogs

Wikipedia

Applications of Quadratic Functions in Algebra (IB Mathematics Analysis and Approaches Standard Level)

Quadratic Functions in Algebra (IB Mathematics Analysis and Approaches Standard Level)

Quadratic functions, typically expressed in the form $y=ax^2+bx+c$ , have numerous practical applications in various fields, making them an essential concept in algebra. Here are some key applications:

Quadratic Functions in Algebra (IB Mathematics Analysis and Approaches Standard Level)

1. Physics and Engineering

Projectile Motion: Quadratic functions describe the path of a projectile under the influence of gravity. The equation $y=ax^2+bx+c$ represents the trajectory of an object, such as a ball being thrown or a rocket being launched.
Quadratic Functions in Algebra (IB Mathematics Analysis and Approaches Standard Level)

Structural Engineering: Quadratic functions are used to model the shapes of parabolic arches and bridges, optimizing material use and ensuring structural stability.
Quadratic Functions in Algebra (IB Mathematics Analysis and Approaches Standard Level)
- Example: The St. Louis Gateway Arch, which has a parabolic shape, can be modeled and analyzed using quadratic functions.
  Quadratic Functions in Algebra (IB Mathematics Analysis and Approaches Standard Level)

2. Economics

Revenue and Profit Maximization: Quadratic functions are used to model revenue and profit functions. The vertex of the parabola indicates the maximum or minimum point, which is crucial for determining optimal pricing strategies.
Quadratic Functions in Algebra (IB Mathematics Analysis and Approaches Standard Level)
- Example: If the revenue $R$ from selling $x$ units of a product is given by $R(x)=ax^2+bx+c$ , finding the vertex helps in determining the quantity that maximizes revenue.
  Quadratic Functions in Algebra (IB Mathematics Analysis and Approaches Standard Level)

Biology

Population Growth: Quadratic functions can model the growth of populations under certain conditions, such as limited resources, where the growth rate eventually decreases as the population size approaches the carrying capacity.
Quadratic Functions in Algebra (IB Mathematics Analysis and Approaches Standard Level)
- Example: The logistic growth model in its simplest form can be approximated by a quadratic function in its early stages.
  Quadratic Functions in Algebra (IB Mathematics Analysis and Approaches Standard Level)

4. Finance

Portfolio Optimization: Quadratic functions are used in mean-variance optimization to determine the portfolio with the maximum expected return for a given level of risk.
Quadratic Functions in Algebra (IB Mathematics Analysis and Approaches Standard Level)
- Example: The quadratic function in the context of finance can represent the trade-off between risk (variance) and return in an investment portfolio.
  Quadratic Functions in Algebra (IB Mathematics Analysis and Approaches Standard Level)

5. Computer Graphics

Bezier Curves: Quadratic Bezier curves are used in computer graphics and animation to create smooth curves. These curves are defined by a quadratic function and are essential for designing scalable vector graphics and animations.
Quadratic Functions in Algebra (IB Mathematics Analysis and Approaches Standard Level)

Quadratic Functions in Algebra (IB Mathematics Analysis and Approaches Standard Level)

A quadratic function is a polynomial function of degree 2. The general form of a quadratic function is:

where: • 𝑎, 𝑏, and 𝑐 are constants, • 𝑎≠0, • 𝑥 is the variable.

Key Features of Quadratic Functions

where (ℎ,𝑘)is the vertex of the parabola. The vertex can also be found using the formula:

Substituting ℎ into the function to find 𝑘:

3. Axis of Symmetry: The axis of symmetry is the vertical line that passes through the vertex. It has the equation: x=ℎ where ℎ is the x-coordinate of the vertex.

4. Roots (or Zeros): The roots of the quadratic function are the values of 𝑥 that make 𝑓(𝑥)=0 They can be found using the quadratic formula:

The expression under the square root, b2 - 4ac, is called the discriminant.

5. Discriminant: The discriminant determines the nature of the roots. if b2 - 4ac > 0 , there are two distinct real roots. if b2 - 4ac = 0, there is one real root (a repeated root). if b2 - 4ac < 0, there are no real roots (the roots are complex)

Transformations of Quadratic Functions Transformations can be used to modify the graph of a quadratic function. Common transformations include:

1. Vertical Translation: Adding or subtracting a constant 𝑘 shifts the graph up or down.

2. Horizontal Translation: Adding or subtracting a constant ℎ to 𝑥 shifts the graph left or right.

3. Reflection: Reflecting across the x-axis changes the sign of 𝑎.

4. Vertical Stretch or Compression: Multiplying by a constant 𝑎 changes the width of the parabola. if |a| > 1, the graph is stretched. if 0 < |a| < 1, the graph is compressed.

Solving Quadratic Equations Quadratic equations can be solved using several methods:

1. Factoring: Write the quadratic equation in factored form:

Set each factor to zero and solve for 𝑥.

2. Completing the Square: Rewrite the quadratic equation in vertex form by completing the square. Solve for 𝑥.

3. Quadratic Formula: Use the quadratic formula:

Applications of Quadratic Functions Quadratic functions have numerous applications, including:

1. Projectile Motion: The path of a projectile under gravity is modeled by a quadratic function. Example: y=ax2 + bx +c, represents the height and 𝑥 the horizontal distance.

2. Optimization Problems: Quadratic functions are used in optimization to find maximum or minimum values. Example: Maximizing the area of a rectangle given a fixed perimeter.

3. Economics: Quadratic functions model various economic phenomena such as profit, cost, and revenue functions.

Problem 1: Consider a quadratic function given by the equation f ( x ) = a x 2 + b x + c . You are told that the vertex of the parabola represented by this equation occurs at (-3,4) and it passes through the point (2,-1). Determine the values of a,b, and c.

Solution: Use vertex form of a quadratic equation. A quadratic equation in vertex form is given by: f ( x ) = a ( x − h ) 2 + k

where (h,k) is the vertex of the parabola. Plugging in the vertex (-3,4) for (h,k), we get:

Substitute the point (2,-1) for (x,y) into the vertex form: We know that f(2) = -1. Substitute 2 for x and -1 for y in the modified equation:

Solve for 𝑎

Substitute 𝑎 back into the vertex form to find 𝑏 and 𝑐 Now that we know a = − 1 5 , substitute 𝑎 back into the vertex form and expand it to standard form:

expand and simplify ( x + 3 ) 2

Conclusion: Therefore, the values of 𝑎, 𝑏, and 𝑐 in the standard form a x 2 + b x + c of the given quadratic function are:

Here's the graph of the quadratic function f ( x ) = a x 2 + b x + c . It shows the vertex at (-3,4) and, also includes the point (2,-1) through which the parabola passes. The graph helps visualize the shape and positioning of the parabola relative to these key points.

Problem 2: Consider a quadratic function defined by the equation f ( x ) = a x 2 + b x + c . The vertex of the parabola described by this equation is located at (1,-5) and it intersects the y-axis at (0,3). Determine the values of a,b, and c.

Solution: Use vertex form of a quadratic equation. The vertex form of a quadratic function is: f ( x ) = a ( x − h ) 2 + k

where (h,k) is the vertex. Substitute the vertex (1,-5) for (h,k), the equation becomes:

Use the y-intercept to find a Since the parabola intersects the y-axis at (0,3). we know f (0)=3. Substitute 0 for 𝑥 and 3 for y in the equation:

Substitute 𝑎 back into the vertex form Now that we have a = 8, so substitute it back: f ( x ) = 8 ( x − 1 ) 2 − 5 Expanding this to standard form:

Conclusion: Thus, the coefficients 𝑎a, 𝑏b, and 𝑐c in the standard form a x 2 + b x + c of the given quadratic function are: a = 8 , b = − 16 , c = 3

Let’s Plot the graph for this Quadratic function..

Here's the graph of the quadratic function f ( x ) = 8 x 2 − 16 x + 3 .The graph visually represents the vertex at (1,-5) and the y-intercept at (0,3). This visualization helps in understanding the shape and key features of the parabola based on the problem's conditions.

Problem 3: You are given a quadratic function described by the equation f ( x ) = a x 2 + b x + c . This parabola has its vertex at (2,-7) and it crosses the x-axis at two points, one of which is (5,0). Determine the values of 𝑎, 𝑏, and 𝑐.

Solution: Use the vertex form of a quadratic equation. The vertex form of a quadratic function is: f ( x ) = a ( x − h ) 2 + k

where (h,k) is the vertex. Given the vertex (2,-7), the equation becomes: f ( x ) = a ( x − 2 ) 2 − 7

Use the x-Intercept to find a Since the parabola touches the x-axis at (5,0), we know f (5) = 0. Substitute 5 for x in the equation:

0 = a ( 5 − 2 ) 2 − 7 0 = 9 a − 7 9 a = 7 a = 7 9

Substitute 𝑎 back into the vertex form Now we have a = 7 9 , so substitute it back:

f ( x ) = 7 9 ( x − 2 ) 2 − 7

Expanding this to standard form:

f ( x ) = 7 9 ( x 2 − 4 x + 4 ) − 7 f ( x ) = 7 9 x 2 − 28 9 x + 28 9 − 7 f ( x ) = 7 9 x 2 − 28 9 x − 35 9

Conclusion: Thus, the coefficients 𝑎, 𝑏, and 𝑐 in the standard form a x 2 + b x + c of the given quadratic function are:

a = 7 9 , b = − 28 9 , c = − 35 9

Let's plot the graph for this quadratic function to visualize the vertex and the x-intercept.

Here's the graph of the quadratic function f ( x ) = 7 9 x 2 − 28 9 x − 35 9 . It visually illustrates the vertex at (2,-7) and the x-intercept at (5,0). This graph helps in understanding the properties of the parabola, especially how it interacts with the x-axis and its vertex characteristics.

Problem 4: You are given a quadratic function described by the equation f ( x ) = a x 2 + b x + c . The vertex of the parabola is located at (-2,3), and it crosses the x-axis at points (-5,0) and (1,0). Determine the values of a, b, and c.

Solution: Use the vertex form of a quadratic equation. The vertex form of a quadratic function is: f ( x ) = a ( x − h ) 2 + k where (h,k) is the vertex. Given the vertex (-2,3), the equation becomes: f ( x ) = a ( x + 2 ) 2 + 3

Use the x-axis intercepts. Given the x-axis intercepts at (-5,0) and (1,0), we know f (-5) = 0 and f (1) = 0. Substitute -5 for x and 0 for y into the equation: For x = -5:

Use the x-axis intercepts. Given the x-axis intercepts at (-5,0) and (1,0), we know f (-5) = 0 and f (1) = 0. Substitute -5 for x and 0 for y into the equation: For x = -5:

0 = a ( − 5 + 2 ) 2 + 3 0 = a ( − 3 ) 2 + 3 0 = 9 a + 3 9 a = − 3 a = − 1 3

For x = 1, let’s verify:

0 = − 1 3 ( 1 + 2 ) 2 + 3 0 = − 1 3 ( 1 + 2 ) 2 + 3 0 = − 3 + 3

0 = 0 (This confirms that: a = − 1 3 is correct.)

Convert to standard form Now we have a = − 1 2 , substitute back into the vertex form and expand:

f ( x ) = − 1 3 ( x + 2 ) 2 + 3 f ( x ) = − 1 3 ( x 2 + 4 x + 4 ) + 3 f ( x ) = − 1 3 x 2 − 4 3 x − 4 3 + 3 f ( x ) = − 1 3 x 2 − 4 3 x + 5 3

Conclusion: Thus, the coefficients a,b, and c in the standard form a x 2 + b x + c of the given quadratic function are:

a = − 1 3 , b = − 4 3 , c = 5 3

Let's visualize this with a graph to see the vertex and the x-axis intercepts.

Here's the graph of the quadratic function f ( x ) = − 1 3 x 2 − 4 3 x + 5 3 . It displays the vertex at (-2,3) and the x-axis intercepts at (-5,0) and (1,0). This visualization helps understand how the parabola behaves with its vertex and intersection points clearly marked.

Applications of Quadratic Functions in Algebra (IB Mathematics Analysis and Approaches Standard Level)

1. Physics and Engineering

2. Economics

Biology

4. Finance

5. Computer Graphics

Leave a comment Cancel reply

Recent Posts

Choose Topic

Recent Comments

Heron's formula

Graphs of Inverse Trigonometry Functions

where:
• 𝑎, 𝑏, and 𝑐 are constants,
• 𝑎≠0,
• 𝑥 is the variable.

where (ℎ,𝑘)is the vertex of the parabola.

The vertex can also be found using the formula:

3. Axis of Symmetry:

The axis of symmetry is the vertical line that passes through the vertex.

It has the equation:
x=ℎ
where ℎ is the x-coordinate of the vertex.

4. Roots (or Zeros):

The roots of the quadratic function are the values of 𝑥 that make 𝑓(𝑥)=0

They can be found using the quadratic formula:

The expression under the square root, b² - 4ac, is called the discriminant.

5. Discriminant:

The discriminant determines the nature of the roots.

if b² - 4ac > 0 , there are two distinct real roots.

if b² - 4ac = 0, there is one real root (a repeated root).

if b² - 4ac < 0, there are no real roots (the roots are complex)

Transformations of Quadratic Functions

Transformations can be used to modify the graph of a quadratic function. Common transformations include:

1. Vertical Translation:

Adding or subtracting a constant 𝑘 shifts the graph up or down.

2. Horizontal Translation:

Adding or subtracting a constant ℎ to 𝑥 shifts the graph left or right.

3. Reflection:

Reflecting across the x-axis changes the sign of 𝑎.

4. Vertical Stretch or Compression:

Multiplying by a constant 𝑎 changes the width of the parabola.

if |a| > 1, the graph is stretched.

if 0 < |a| < 1, the graph is compressed.

Solving Quadratic Equations

Quadratic equations can be solved using several methods:

1. Factoring:

Write the quadratic equation in factored form:

2. Completing the Square:

Rewrite the quadratic equation in vertex form by completing the square. Solve for 𝑥.

3. Quadratic Formula:

Use the quadratic formula:

Applications of Quadratic Functions

Quadratic functions have numerous applications, including:

1. Projectile Motion:

The path of a projectile under gravity is modeled by a quadratic function.

Example: y=ax² + bx +c, represents the height and 𝑥 the horizontal distance.

2. Optimization Problems:

Quadratic functions are used in optimization to find maximum or minimum values.

Example: Maximizing the area of a rectangle given a fixed perimeter.

3. Economics:

Quadratic functions model various economic phenomena such as profit, cost, and revenue functions.

Problem 1:

Consider a quadratic function given by the equation $f (x) = a x^{2} + b x + c$ . You are told that the vertex of the parabola represented by this equation occurs at (-3,4) and it passes through the point (2,-1). Determine the values of a,b, and c.

Solution:

Use vertex form of a quadratic equation.
A quadratic equation in vertex form is given by:
$f (x) = a (x - h)^{2} + k$

Substitute the point (2,-1) for (x,y) into the vertex form:
We know that f(2) = -1. Substitute 2 for x and -1 for y in the modified equation:

Substitute 𝑎 back into the vertex form to find 𝑏 and 𝑐

Now that we know $a = - \frac{1}{5}$ , substitute 𝑎 back into the vertex form and expand it to standard form:

expand and simplify $(x + 3)^{2}$

Conclusion:
Therefore, the values of 𝑎, 𝑏, and 𝑐 in the standard form $a x^{2} + b x + c$ of the given quadratic function are:

Here's the graph of the quadratic function $f (x) = a x^{2} + b x + c$ . It shows the vertex at (-3,4) and, also includes the point (2,-1) through which the parabola passes. The graph helps visualize the shape and positioning of the parabola relative to these key points.

Problem 2:

Consider a quadratic function defined by the equation $f (x) = a x^{2} + b x + c$ . The vertex of the parabola described by this equation is located at (1,-5) and it intersects the y-axis at (0,3). Determine the values of a,b, and c.

Solution:

Use vertex form of a quadratic equation.
The vertex form of a quadratic function is:
$f (x) = a (x - h)^{2} + k$

Use the y-intercept to find a

Since the parabola intersects the y-axis at (0,3). we know f (0)=3. Substitute 0 for 𝑥 and 3 for y in the equation:

Substitute 𝑎 back into the vertex form

Now that we have a = 8, so substitute it back:

$f (x) = 8 (x - 1)^{2} - 5$

Expanding this to standard form:

Conclusion:

Thus, the coefficients 𝑎a, 𝑏b, and 𝑐c in the standard form $a x^{2} + b x + c$ of the given quadratic function are:

$a = 8, b = - 16, c = 3$

Here's the graph of the quadratic function $f (x) = 8 x^{2} - 16 x + 3$ .The graph visually represents the vertex at (1,-5)
and the y-intercept at (0,3). This visualization helps in understanding the shape and key features of the parabola based on the problem's conditions.

Problem 3:

You are given a quadratic function described by the equation
$f (x) = a x^{2} + b x + c$ . This parabola has its vertex at (2,-7) and it crosses the x-axis at two points, one of which is (5,0). Determine the values of 𝑎, 𝑏, and 𝑐.

Solution:

Use the vertex form of a quadratic equation.
The vertex form of a quadratic function is:
$f (x) = a (x - h)^{2} + k$

where (h,k) is the vertex. Given the vertex (2,-7), the equation becomes:
$f (x) = a (x - 2)^{2} - 7$

Use the x-Intercept to find a
Since the parabola touches the x-axis at (5,0), we know f (5) = 0. Substitute 5 for x in the equation:

$\begin{aligned} 0 = a (5 - 2)^{2} - 7 \\ 0 = 9 a - 7 \\ 9 a = 7 \\ a = \frac{7}{9} \end{aligned}$

Substitute 𝑎 back into the vertex form
Now we have $a = \frac{7}{9}$ , so substitute it back:

$f (x) = \frac{7}{9} (x - 2)^{2} - 7$

$\begin{aligned} f (x) = \frac{7}{9} (x^{2} - 4 x + 4) - 7 \\ f (x) = \frac{7}{9} x^{2} - \frac{28}{9} x + \frac{28}{9} - 7 \\ f (x) = \frac{7}{9} x^{2} - \frac{28}{9} x - \frac{35}{9} \end{aligned}$

Conclusion:
Thus, the coefficients 𝑎, 𝑏, and 𝑐 in the standard form $a x^{2} + b x + c$ of the given quadratic function are:

$a = \frac{7}{9}, b = - \frac{28}{9}, c = - \frac{35}{9}$

Problem 4:
You are given a quadratic function described by the equation $f (x) = a x^{2} + b x + c$ .
The vertex of the parabola is located at (-2,3), and it crosses the x-axis at points (-5,0) and (1,0). Determine the values of a, b, and c.

Solution:

Use the vertex form of a quadratic equation.
The vertex form of a quadratic function is:
$f (x) = a (x - h)^{2} + k$

where (h,k) is the vertex. Given the vertex (-2,3), the equation becomes:
$f (x) = a (x + 2)^{2} + 3$

Use the x-axis intercepts.

Given the x-axis intercepts at (-5,0) and (1,0), we know f (-5) = 0 and f (1) = 0.
Substitute -5 for x and 0 for y into the equation:

For x = -5:

Use the x-axis intercepts.

Given the x-axis intercepts at (-5,0) and (1,0), we know f (-5) = 0 and f (1) = 0.
Substitute -5 for x and 0 for y into the equation:

For x = -5:

$\begin{aligned} 0 = a (- 5 + 2)^{2} + 3 \\ 0 = a (- 3)^{2} + 3 \\ 0 = 9 a + 3 \\ 9 a = - 3 \\ a = - \frac{1}{3} \end{aligned}$

$\begin{matrix} 0 = - \frac{1}{3} (1 + 2)^{2} + 3 \\ 0 = - \frac{1}{3} (1 + 2)^{2} + 3 \\ 0 = - 3 + 3 \end{matrix}$

$0 = 0 (This confirms that: a = - \frac{1}{3} is correct.)$

Convert to standard form

Now we have $a = \frac{- 1}{2}$ , substitute back into the vertex form and expand:

$\begin{aligned} f (x) = - \frac{1}{3} (x + 2)^{2} + 3 \\ f (x) = - \frac{1}{3} (x^{2} + 4 x + 4) + 3 \\ f (x) = - \frac{1}{3} x^{2} - \frac{4}{3} x - \frac{4}{3} + 3 \\ f (x) = - \frac{1}{3} x^{2} - \frac{4}{3} x + \frac{5}{3} \end{aligned}$

Conclusion:

Thus, the coefficients a,b, and c in the standard form $a x^{2} + b x + c$ of the given quadratic function are:

$a = - \frac{1}{3}, b = - \frac{4}{3}, c = \frac{5}{3}$