Chapter 1.1, Problem 54E

Solution

For the graph of $y=a{x}^2+bx+c$ , show that the y-coordinate of the vertex is $\frac{-b^2}{4a}+c$ .

Proofed.

  $y=\frac{-b^2}{4a}+c$ .

Given:

The given function $y=a{x}^2+bx+c$

Concept Used:

• The vertex is the highest point if the parabola opens downward and the lowest point if the parabola opens upward.
• The axis of symmetry is the line that cuts the parabola into 2 matching halves and the vertex lies on the axis of symmetry.

Calculation:

The given function $y=a{x}^2+bx+c$

Here $a=a,b=b,c=c$

Axis of symmetry (x) $=\frac{-b}{2a}$

So, the y-coordinate of the vertex is $=y\left(\frac{-b}{2a}\right)$

  $\begin{array}{c}y\left(\frac{-b}{2a}\right)=a{x}^2+bx+c\\ =a{\left(\frac{-b}{2a}\right)}^2+b\left(\frac{-b}{2a}\right)+c\\ =a\frac{b^2}{4a^2}-\frac{b}{2a}+c\\ =\frac{b^2}{4a}-\frac{b}{2a}+c\\ =\frac{b^2-2b^2}{4a}+c\\ =-\frac{b^2}{4a}+c\end{array}$

So, the y-coordinate of the vertex is $-\frac{b^2}{4a}+c$ .

Hence proofed.