Quadratic Equations

A quadratic function is a function that can be written in the form f(x) = a(x − h)² + k, where a ≠ 0. The U-shaped graph of a quadratic function is called a parabola.

 

EXAMPLE: \n ​​Describe the transformation of f(x) = x² represented by g(x) = (x + 4)² − 1. Then graph each function.

SOLUTION:

Notice that the function is of the form g(x) = (x − h)² + k. Rewrite the function to identify h and k.

 

Because h = −4 and k = −1, the graph of g is a translation 4 units left and 1 unit down of the graph of f.

 

EXAMPLE:

Describe the transformation of f(x) = x² represented by g. Then graph g(x) = −1/2x²

SOLUTION:

Notice that the function is of the form g(x) = −ax², where a = 1/2 . So, the graph of g is a reflection in the x-axis and a vertical shrink by a factor of 1/2 of the graph of f

 

The lowest point on a parabola that opens up or the highest point on a parabola that opens down is the vertex. The vertex form of a quadratic function is f(x) = a(x − h)² + k, where a ≠ 0 and the vertex is (h, k).

 

EXAMPLE:

Let the graph of g be a vertical stretch by a factor of 2 and a reflection in the x-axis, followed by a translation 3 units down of the graph of f(x) = x². Write a rule for g and identify the vertex.

SOLUTION:

Identify how the transformations affect the constants in vertex form.

reflection in x-axis a = −2

vertical stretch by 2

translation 3 units down k = −3

Write the transformed function. g(x) = a(x − h)² + k

Vertex form of a quadratic function = −2(x − 0)² + (−3)

Substitute −2 for a, 0 for h, and −3 for  = −2x² − 3

The transformed function is g(x) = −2x² − 3. The vertex is (0, −3)

EXAMPLE:

Let the graph of g be a translation 3 units right and 2 units up, followed by a reflection in the y-axis of the graph of f(x) = x² − 5x. Write a rule for g.

SOLUTION:

First write a function h that represents the translation of f

 

Then write a function g that represents the reflection of h.

 

An axis of symmetry is a line that divides a parabola into mirror images and passes through the vertex. Because the vertex of f(x) = a(x − h)² + k is (h, k), the axis of symmetry is the vertical line x = h.

 

EXAMPLE: \n Graph f(x) = −2(x + 3)² + 4. Label the vertex and axis of symmetry.

SOLUTION: \n Identify the constants a = −2, h = −3, and k = 4.

Plot the vertex (h, k) = (−3, 4) and draw the axis of symmetry x = −3. Evaluate the function for two values of x

Plot the points (−2, 2), (−1, −4), and their reflections in the axis of symmetry.

Draw a parabola through the plotted points.

 

Quadratic functions can also be written in standard form, f(x) = ax² + bx + c, where a ≠ 0. You can derive standard form by expanding vertex form.

 

 

EXAMPLE:

Graph f(x) = 3x² − 6x + 1. Label the vertex and axis of symmetry.

SOLUTION: \n Identify the coefficients a = 3, b = −6, and c = 1. Because a > 0, the parabola opens up.

Find the vertex. First calculate the x-coordinate. Then find the y-coordinate of the vertex. f(1) = 3(1)² − 6(1) + 1 = −2 So, the vertex is (1, −2). Plot this point.

Draw the axis of symmetry x = 1.

Identify the y-intercept c, which is 1. Plot the point (0, 1) and its reflection in the axis of symmetry, (2, 1).

Evaluate the function for another value of x, such as x = 3. f(3) = 3(3)² − 6(3) + 1 = 10 Plot the point (3, 10) and its refl ection in the axis of symmetry, (−1, 10).

Draw a parabola through the plotted points.

 

Because the vertex is the highest or lowest point on a parabola, its y-coordinate is the maximum value or minimum value of the function. The vertex lies on the axis of symmetry, so the function is increasing on one side of the axis of symmetry and decreasing on the other side.

 

EXAMPLE:

Find the minimum value or maximum value of f(x) = 1/2 x² − 2x − 1. Describe the domain and range of the function, and where the function is increasing and decreasing.

SOLUTION: \n Identify the coefficients a = 1/2 , b = −2, and c = −1. Because a > 0, the parabola opens up and the function has a minimum value. To find the minimum value, calculate the coordinates of the vertex.

 

The minimum value is −3. So, the domain is all real numbers and the range is y ≥ −3. The function is decreasing to the left of x = 2 and increasing to the right of x = 2.

When the graph of a quadratic function has at least one x-intercept, the function can be written in intercept form, f(x) = a(x − p)(x − q), where a ≠ 0.

 

EXAMPLE:

Graph f(x) = −2(x + 3)(x − 1). Label the x-intercepts, vertex, and axis of symmetry

SOLUTION:

Identify the x-intercepts. The x-intercepts are p = −3 and q = 1, so the parabola passes through the points (−3, 0) and (1, 0)

Find the coordinates of the vertex.

So, the axis of symmetry is x = −1 and the vertex is (−1, 8).

Draw a parabola through the vertex and the points where the x-intercepts occur.

 

Previously, you learned that the graph of a quadratic function is a parabola that opens up or down. A parabola can also be defined as the set of all points (x, y) in a plane that are equidistant from a fixed point called the focus and a fixed line called the directrix.

 

EXAMPLE:

Use the Distance Formula to write an equation of the parabola with focus F(0, 2) and directrix y = −2

SOLUTION:

Notice the line segments drawn from point F to point P and from point P to point D. By the definition of a parabola, these line segments must be congruent.

 

 

EXAMPLE: \n Write an equation of the parabola shown

 

SOLUTION:

Because the vertex is not at the origin and the axis of symmetry is horizontal, the equation has the form x = 1/4p (y − k)2 + h. The vertex (h, k) is (6, 2) and the focus (h + p, k) is (10, 2), so h = 6, k = 2, and p = 4. Substitute these values to write an equation of the parabola.

So, an equation of the parabola is x = 1/16 (y − 2)² + 6.