Chapter 1.2, Problem 16E

Solution

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

Axis of symmetry (x) $=\frac{-b}{2a}=3$

Vertex $=(3,-8)$

X-intercept = $1,5$

Given:

The given function $f(x)=2(x-5)(x-1)$

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 $f(x)=2(x-5)(x-1)$ is -

  

The given function $f(x)=2(x-5)(x-1)$

  $\begin{array}{c}f(x)=2(x-5)(x-1)\\ f(x)=2(x\cdot x+x\cdot (-1)-5\cdot x-5\cdot (-1))\\ =2(x^2-x-5x+5)\\ =2(x^2-6x+5)\\ =2x^2-12x+10\end{array}$

Here $a=2,b=-12,c=10$

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

  $\begin{array}{l}=\frac{-(-12)}{2(2)}\\ =\frac{12}{4}\\ =3\end{array}$

So Axis of symmetry (x) $=\frac{-b}{2a}=3$

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

  $\begin{array}{c}f\left(\frac{-b}{2a}\right)=f\left(3\right)\\ f(x)=2(x-5)(x-1)\\ f(3)=2(3-5)(3-1)\\ =2(-2)(2)\\ =-8\end{array}$

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

Vertex $=(3,-8)$

x-intercept- $f(x)=2x^2-12x+10$

  $\begin{array}{c}f(x)=2x^2-12x+10\\ forx-\mathrm{int}erceptf(x)=0\\ 0=2x^2-12x+10\\ 2x^2-12x+10=0\\ x^2-6x+5=0\\ x^2-(5+1)x+5=0\\ x^2-5x-x+5=0\\ x(x-5)-1(x-5)=0\\ (x-1)(x-5)=0\\ (x-1)=0\\ x=1\\ (x-5)=0\\ x=5\\ x=1,5\end{array}$

X-intercept $1,5$