By T. Barron & S. Kastberg

## Lesson 7

The quadratic formula is used to solve for x in a quadratic function. It can be derived from using the general form of the quadratic equation, y = a x2 + bx + c, by setting the equation = 0 and using the "completing the square" technique that we used in the previous less to solve for x.

If we perform these calculations, we end up with a formula that solves any quadratic equation for x.

where "a" is the coefficient of the x2 term, "b" is the coefficient of the x term, and c is the constant.

So, why do we need to know the quadratic formula? Well, in a previous lesson, when we were trying to find the x-intercepts, we could only approximate them because we didn't know how to solve the equation, y = -3x2+ x + 1. We said the the x-intercepts were approximately x = -0.8 and x = 0.4. , but we couldn't tell for sure.

So, let's use quadratic formula to solve for the x-intercepts.

To find the x-intercepts of any equation, substitute 0 in for y and solve for x.

So, we have 0 = 3x2+ x + 1. Now, use the quadratic equation to solve for x, wich a = 3, b = 1, and c = 1:

So, now we can find the value of the x-intercepts and not have to estimate!

What about if there are no x-intercepts?

Consider the quadratic function: f(x) = x2 - 3x + 5.

Here, a = 1, b = -3 and c = 5

Let's find the y-intercept first:

To find the y-intercept, substitute 0 in for x and solve for f(x):

f(0) = (0)2 - 3(0) + 5.

f(0) = 5

So the y-intercept is (0, 5).

Now, let's find the vertex:

To find the x-value of the vertex, use the formula x = -b/2a.

So,

To find the y-value of the vertex, substitute 3/2 in for x in the equation , and solve for f(x):

So, the vertex is (3/2, 11/4)

Finally, let's find the x-intercepts by using the quadratic formula.

To do this, we set the y-value (or f(x)) = 0 and solve for x. So, take the function , and set it equal to 0:

Now use the quadratic formula where a = 1, b = -3, c = 5

Well,is not a real number. If you try to "plug" into your calculator, you will get an error. So, if we can't solve for x, that means there are no x-intercepts. Let's graph the parabola using the y-intercept (0, 5) and the vertex (3/2, 11/4). Remember, the parabola should not cross the x-axis anywhere. Also, remember since "a" is positive, the graph should open upward. Let's see!

Graph of f(x) = x2 - 3x + 5.

So, as we can see, being able to use the quadratic formula is quite important when graphing quadratic functions!

Graph the quadratic functions below by finding the y-intercept, the x-intercept(s) and the vertex of each function.

1. f(x) = - x2 - 4x + 1

2. f(x) = 2 (x + 1)2 - 7

3. f(x) = -5x2 + 2x - 1

4. f(x) = x2 - x

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

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