Algebra Know-It-ALL

410 Graphs of Quadratic Functions

Are you confused?

We’ve made a substantial claim: The x-value of the extremum for a quadratic function

y=ax^2 +bx+c

is equal to −b/(2a) when there are no real zeros. This holds true whether the parabola opens upward or

downward. It’s also true for quadratic functions that have one or two real zeros. Do you wonder how we

know this? I haven’t proven it or even derived it in a general way.

This fact comes out of a maneuver that requires differential calculus. We won’t get into calculus in this

book, but the approach can be explained qualitatively. It involves finding the point on the parabola where

a line tangent to the curve has a slope of 0. That’s the point where the curve is “locally horizontal,” and

it’s always the vertex in a quadratic function with real coefficients and a real constant. If you look at all the

graphs in this chapter, you’ll see that a straight line drawn tangent to the curve, and passing through the

vertex, is always horizontal. Differential calculus gives us a formula for the slope m of a quadratic function

at any point based on the x-value at that point:

m= 2 ax+b

If we set m= 0 to make the slope horizontal, we get

0 = 2 ax+b

That’s a first-degree equation in x. When we solve it, we get

x=−b/(2a)

Here’s a challenge!

Consider the following quadratic function with no real zeros:

y= 2 x^2 + 4 x+ 3

Find out whether the parabola for this function opens upward or downward. After that, find the x-value

and the y-value of the extremum. Then locate two other points on the curve, and draw an approximation

of the parabola representing the function.

Solution

This parabola opens upward, because the coefficient of x^2 is positive. That means it has an absolute mini-

mum. Remember again the general polynomial form for a quadratic function:

y=ax^2 +bx+c

Thex-value at the absolute minimum point is

xmin=−b/(2a)

=−4 / (2 × 2)

= (−4)/4

=− 1