In this video, we are going to look at how to evaluate quadratic functions. Quadratic functions are written in the form y=ax^2+bx+c or f(x)=ax^2+bx+c. After you finish this lesson, view all of our Algebra 1 lessons and practice problems.

For example:
When given y=3x^2+2x-5 at x=2, we substitute 2 in for where we see x.
y=3(2)^2+2(2)-5
Remember to always plug in values in parenthesis (to avoid confusion with signs). Then, this can be simplified to
y=3(4)+2(2)-5
and even further to
y=12+4-5
to get a final answer of
y=11

Example of Evaluating Quadratic Functions

Example 1

y=x^2-7x-13 where x = 6

So let’s substitute.

y=(6)^2-7(6)-13
y=36-42-13
y=-19

Example 2

f(x) = 4b^2+6b-4 ; find f(5)
So let’s substitute.
f(5)=4(5)^2+6(5)-4
f(5)=4(25)+6(5)-4
f(5)=100+35-4
f(5)=135-4
f(5)=131

Video-Lesson Transcript

Let’s go over how to evaluate quadratic functions.

Quadratic functions have this form:

y = ax^2 + bx + c

The most important thing to note here is the presence of x^2. This is the highest exponent.

Or this form:

f(x) = ax^2 + bx + c

One quadratic function could be

y = 3x^2 + 2x - 5 where x = 2

So let’s substitute.

y = 3 (2)^2 + 2 (2) - 5
y = 3 (4) + 2 (2) - 5
y = 12 + 4 - 5
y = 11
Evaluating Quadratic Functions

Let’s look at another example in a different form.

f(x) = 2x^2 - 4x + 7, f(-3)

Now, let’s substitute.

f(x) = 2x^2 - 4x + 7
f(-3) = 2 (-3)^2 - 4 (-3) + 7
f(-3) = 2 (9) - 4 (-3) + 7
f(-3) = 18 + 12 + 7
f(-3) = 37