Chapter 1.2, Problem 17E

Solution

Graph the function. Label the vertex and axis of symmetry and x-intercepts.

Axis of symmetry (x) $=\frac{-b}{2a}=-1$

Vertex $=(-1,25)$

X-intercept = $4,-6$

Given:

The given function $y=-(x-4)(x+6)$

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.
• The x-intercept is the point where a line crosses the x-axis.

Calculation:

The graph of the given function $y=-(x-4)(x+6)$ is -

  

The given function $y=-(x-4)(x+6)$

  $\begin{array}{c}y=-(x-4)(x+6)\\ y=-(x\cdot x+x\cdot (6)-4\cdot x-4\cdot (6))\\ y=-(x^2+6x-4x-24)\\ y=-(x^2+2x-24)\\ y=-x^2-2x+24\end{array}$

Here $a=-1,b=2,c=24$

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

  $\begin{array}{l}=\frac{-(-2)}{2(-1)}\\ =\frac{2}{-2}\\ =-1\end{array}$

So Axis of symmetry (x) $=\frac{-b}{2a}=-1$

Vertex $=\left(\frac{-b}{2a},y(\frac{-b}{2a})\right)$

  $\begin{array}{c}y\left(\frac{-b}{2a}\right)=y\left(-1\right)\\ y=-(x-4)(x+6)\\ y=-(-1-4)(-1+6)\\ y=-(-5)(5)\\ y=25\end{array}$

Vertex $=\left(\frac{-b}{2a},y(\frac{-b}{2a})\right)$

Vertex $=(-1,25)$

x-intercept- $y=-x^2-2x+24$

  $\begin{array}{c}y=-x^2-2x+24\\ forx-\mathrm{int}ercepty=0\\ 0=-x^2-2x+24\\ -x^2-2x+24=0\\ x^2+2x-24=0\\ x^2+(6-4)x-24=0\\ x^2+6x-4x-24=0\\ x(x+6)-4(x+6)=0\\ (x-4)(x+6)=0\\ (x-4)=0\\ x=4\\ (x+6)=0\\ x=-6\\ x=4,-6\end{array}$

X-intercept = $4,-6$