Lesson Objectives

- Demonstrate an understanding of parabolas
- Learn how to find the vertex of a parabola in standard form
- Learn how to find the vertex using the vertex formula
- Learn how to use the discriminant to determine the number of x-intercepts
- Learn how to sketch the graph of any parabola

## How to Find the Vertex of a Parabola in Standard Form

In the last lesson, we introduced the graph of a quadratic equation, which is known as a "parabola". In that lesson, we looked at the simplest quadratic function,
known as the "squaring function".

f(x) = x

We learned how to graph related parabolas based on horizontal and vertical shifts. Additionally, we learned about the vertex. The vertex of a parabola is the lowest point for an upward facing parabola or the highest point for a downward facing parabola. When our quadratic equation is in vertex form, the vertex is given as the point (h,k).

Vertex Form:

f(x) = a(x - h)

In most cases, our quadratic equation is not in vertex form. This means we need to do some additional work to find our vertex. There are two methods we can use. First and most tedious, we can use a method that involves completing the square. The second method, which is much easier, is a method that uses a formula known as the "vertex formula". The vertex formula is derived from the completing the square method. Let's take a look at an example where we find our vertex from completing the square.

Example 1: Find the vertex by completing the square.

f(x) = x

We want to place this equation in the format of:

f(x) = a(x - h)

f(x) = x

To complete the square, add 1/2 of the coefficient on the first-degree term squared to the right side of the equation. To make this legal, we will also subtract it away. The process may seem a bit confusing at first, but watch this example in action:

The coefficient of our first-degree term is 4, if we take 1/2 of this and square it, we get 4. This means we will add and subtract 4 away from the right side of the equation.

f(x) = x

f(x) = (x

We can factor the perfect square trinomial:

f(x) = (x + 2)

Now we have our desired form, we can rewrite our function as:

f(x) = a(x - h)

f(x) = (x - (-2))

Since our vertex occurs at (h,k), we can state our vertex occurs at (-2, -1).

Example 2: Find the vertex by completing the square.

f(x) = -2x

When we complete the square, the coefficient of the squared variable needs to be 1. Let's factor out a -2 from the variable terms on the left side of the equation:

f(x) = -2(x

Now we can complete the square. The coefficient of the first-degree term is -6. If we multiply this by one-half and square the result, we will get 9. Let's add 9 and subtract 9 from the left side of the equation. Be careful, we have to do this inside of the parentheses.

f(x) = -2(x

Now we have this extra -9 inside of the parentheses, we can't just move it outside. We need to first multiply it by -2, then it can be moved outside of the parentheses:

f(x) = -2(x

Now we can factor the perfect square trinomial:

f(x) = -2(x - 3)

Our vertex occurs at (3, 1)

Example 3: find the vertex using the vertex formula. $$f(x) = x^2 - 6x + 11$$ First, let's record the values for a and b:

a, the coefficient of x

Let's plug into the vertex formula: $$\left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)$$ $$\left(\frac{-(-6)}{2(1)}, f\left(\frac{-b}{2a}\right)\right)$$ $$(3, f(3))$$ We need to plug a 3 in for x in our equation: $$f(3) = (3)^2 - 6(3) + 11$$ $$f(3) = 9 - 18 + 11 = 2$$ Our vertex occurs at the point (3,2).

When the discriminant is greater than 0, we have two real solutions. $$b^2 - 4ac > 0$$ When the discriminant is equal to 0, we have one real solution. $$b^2 - 4ac = 0$$ When the discriminant is less than 0, we have two imaginary solutions. $$b^2 - 4ac < 0$$ When we think about the x-intercept or intercepts, these will occur when y is 0. $$y = ax^2 + bx + c$$ $$0 = ax^2 + bx + c$$ $$ax^2 + bx + c = 0$$ Since we know this equation can be solved using the quadratic formula, we know the x-intercepts follow the same pattern. When the discriminant is larger than 0, we will have two x-intercepts, when the discriminant is equal to 0, we will have only one. In the case, where the discriminant is less than 0, we will not have an x-intercept. Let's look at an example.

Example 4: Determine the number of x-intercepts $$f(x) = -2x^2 + 4x - 6$$ We record the values for a, b, and c.

a = -2, b = 4, c = -6

Plug into the discriminant: $$b^2 - 4ac$$ $$(4)^2 - 4(-2)(-6) = 16 - 48 = -32$$ Since the discriminant is less than 0, we will not have any x-intercepts.

Example 5: Graph each $$f(x) = -x^2 + 6x - 8$$ Since a, the coefficient of the squared variable is -1, we know the parabola opens down. Let's find the vertex.

Vertex: $$\left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)$$ $$\left(\frac{-6}{2(-1)}, f\left(\frac{-b}{2a}\right)\right)$$ $$(3, f(3))$$ $$f(3) = -(3)^2 + 6(3) - 8$$ $$f(3) = -9 + 18 - 8 = 1$$ Our vertex will occur a the point (3, 1).

Let's think about the x-intercepts: $$-x^2 + 6x - 8 = 0$$ $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$x = \frac{-6 \pm \sqrt{6^2 - 4(-1)(-8)}}{2(-1)}$$ $$x = \frac{-6 \pm 2}{-2}$$ $$x = 2, 4$$ Our x-intercepts will occur at (4,0) and (2,0). Let's think about the y-intercept. This occurs when x is 0. $$f(x) = -x^2 + 6x - 8$$ $$f(0) = -(0)^2 + 6(0) - 8 = -8$$ Our y-intercept will occur at (0,-8). Since a parabola is symmetric about its axis, we can determine that another point will occur at (6,-8). We now have five points which can be used to sketch our graph.

f(x) = x

^{2}We learned how to graph related parabolas based on horizontal and vertical shifts. Additionally, we learned about the vertex. The vertex of a parabola is the lowest point for an upward facing parabola or the highest point for a downward facing parabola. When our quadratic equation is in vertex form, the vertex is given as the point (h,k).

Vertex Form:

f(x) = a(x - h)

^{2}+ kIn most cases, our quadratic equation is not in vertex form. This means we need to do some additional work to find our vertex. There are two methods we can use. First and most tedious, we can use a method that involves completing the square. The second method, which is much easier, is a method that uses a formula known as the "vertex formula". The vertex formula is derived from the completing the square method. Let's take a look at an example where we find our vertex from completing the square.

Example 1: Find the vertex by completing the square.

f(x) = x

^{2}+ 4x + 3We want to place this equation in the format of:

f(x) = a(x - h)

^{2}+ kf(x) = x

^{2}+ 4x + 3To complete the square, add 1/2 of the coefficient on the first-degree term squared to the right side of the equation. To make this legal, we will also subtract it away. The process may seem a bit confusing at first, but watch this example in action:

The coefficient of our first-degree term is 4, if we take 1/2 of this and square it, we get 4. This means we will add and subtract 4 away from the right side of the equation.

f(x) = x

^{2}+ 4x + 4 - 4 + 3f(x) = (x

^{2}+ 4x + 4) - 4 + 3We can factor the perfect square trinomial:

f(x) = (x + 2)

^{2}- 1Now we have our desired form, we can rewrite our function as:

f(x) = a(x - h)

^{2}+ kf(x) = (x - (-2))

^{2}+ (-1)Since our vertex occurs at (h,k), we can state our vertex occurs at (-2, -1).

Example 2: Find the vertex by completing the square.

f(x) = -2x

^{2}+ 12x - 17When we complete the square, the coefficient of the squared variable needs to be 1. Let's factor out a -2 from the variable terms on the left side of the equation:

f(x) = -2(x

^{2}- 6x) - 17Now we can complete the square. The coefficient of the first-degree term is -6. If we multiply this by one-half and square the result, we will get 9. Let's add 9 and subtract 9 from the left side of the equation. Be careful, we have to do this inside of the parentheses.

f(x) = -2(x

^{2}- 6x + 9 - 9) - 17Now we have this extra -9 inside of the parentheses, we can't just move it outside. We need to first multiply it by -2, then it can be moved outside of the parentheses:

f(x) = -2(x

^{2}- 6x + 9) + 18 - 17Now we can factor the perfect square trinomial:

f(x) = -2(x - 3)

^{2}+ 1Our vertex occurs at (3, 1)

### Vertex Formula

A shortcut to the method where we find the vertex by completing the square is known as the vertex formula. This formula is derived from the completing the square method. We will show the steps here, but it is not important to memorize this part. We only need to know the formula, which is given at the end. $$f(x) = ax^2 + bx + c$$ Factor out the a from the variable terms: $$f(x) = a\left(x^2 + \frac{b}{a}x\right) + c$$ Complete the square: $$f(x) = a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2} \right) + c$$ Distribute the a to the final term inside of the parentheses, and then remove it from the parentheses. We can then factor the perfect square trinomial. $$f(x) = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c$$ Let's get a common denominator and change the format: $$f(x) = a\left(x - \left(-\frac{b}{2a}\right)\right)^2 + \frac{4ac - b^2}{4a}$$ For the vertex, the x-coordinate is: $$\frac{-b}{2a}$$ The y-coordinate is: $$\frac{4ac - b^2}{4a}$$ To make it easier, we can write the y-coordinate as: $$f\left(\frac{-b}{2a}\right)$$ After all that messy math, we can state that a quadratic equation of the form: $$f(x) = ax^2 + bx + c, a≠ 0$$ has a vertex that occurs at the point: $$\left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)$$ Let's look at an example.Example 3: find the vertex using the vertex formula. $$f(x) = x^2 - 6x + 11$$ First, let's record the values for a and b:

a, the coefficient of x

^{2}is 1, while b, the coefficient for x, is -6.Let's plug into the vertex formula: $$\left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)$$ $$\left(\frac{-(-6)}{2(1)}, f\left(\frac{-b}{2a}\right)\right)$$ $$(3, f(3))$$ We need to plug a 3 in for x in our equation: $$f(3) = (3)^2 - 6(3) + 11$$ $$f(3) = 9 - 18 + 11 = 2$$ Our vertex occurs at the point (3,2).

### Using the Discriminant to Determine the Number of X-Intercepts

A few lessons ago, we learned about the quadratic formula. If a quadratic equation is in standard form, we can record the values for a, b, and c and use the formula to obtain a solution. $$ax^2 + bx + c = 0$$ $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ The part under the square root symbol is known as the discriminant. $$b^2 - 4ac$$ At this point, we should know the following:When the discriminant is greater than 0, we have two real solutions. $$b^2 - 4ac > 0$$ When the discriminant is equal to 0, we have one real solution. $$b^2 - 4ac = 0$$ When the discriminant is less than 0, we have two imaginary solutions. $$b^2 - 4ac < 0$$ When we think about the x-intercept or intercepts, these will occur when y is 0. $$y = ax^2 + bx + c$$ $$0 = ax^2 + bx + c$$ $$ax^2 + bx + c = 0$$ Since we know this equation can be solved using the quadratic formula, we know the x-intercepts follow the same pattern. When the discriminant is larger than 0, we will have two x-intercepts, when the discriminant is equal to 0, we will have only one. In the case, where the discriminant is less than 0, we will not have an x-intercept. Let's look at an example.

Example 4: Determine the number of x-intercepts $$f(x) = -2x^2 + 4x - 6$$ We record the values for a, b, and c.

a = -2, b = 4, c = -6

Plug into the discriminant: $$b^2 - 4ac$$ $$(4)^2 - 4(-2)(-6) = 16 - 48 = -32$$ Since the discriminant is less than 0, we will not have any x-intercepts.

### Graphing a Parabola

We can put all of the information from this lesson together and come up with a procedure to graph a quadratic function of the form: $$f(x) = ax^2 + bx + c$$- Determine if the graph opens up or down
- If a > 0, the parabola opens up
- If a < 0, the parabola opens down

- Locate and plot the vertex using the vertex formula
- Plot the intercepts
- y-intercept occurs at f(0)
- x-intercept occurs at f(x) = 0

- Plot additional points as needed and sketch the graph

Example 5: Graph each $$f(x) = -x^2 + 6x - 8$$ Since a, the coefficient of the squared variable is -1, we know the parabola opens down. Let's find the vertex.

Vertex: $$\left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)$$ $$\left(\frac{-6}{2(-1)}, f\left(\frac{-b}{2a}\right)\right)$$ $$(3, f(3))$$ $$f(3) = -(3)^2 + 6(3) - 8$$ $$f(3) = -9 + 18 - 8 = 1$$ Our vertex will occur a the point (3, 1).

Let's think about the x-intercepts: $$-x^2 + 6x - 8 = 0$$ $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$x = \frac{-6 \pm \sqrt{6^2 - 4(-1)(-8)}}{2(-1)}$$ $$x = \frac{-6 \pm 2}{-2}$$ $$x = 2, 4$$ Our x-intercepts will occur at (4,0) and (2,0). Let's think about the y-intercept. This occurs when x is 0. $$f(x) = -x^2 + 6x - 8$$ $$f(0) = -(0)^2 + 6(0) - 8 = -8$$ Our y-intercept will occur at (0,-8). Since a parabola is symmetric about its axis, we can determine that another point will occur at (6,-8). We now have five points which can be used to sketch our graph.

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