Question

Please help me on number one It’s all one question

Answer 1

Given:

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

a) Standard form of the function is,

$\begin{gathered} f(x)=2x^2+12x+10 \\ f(x)=2(x^2+6x+5+9-9) \\ f(x)=2(x^2+6x+9-4) \\ f(x)=2(x+3)^2-8 \end{gathered}$

Standard form is,

$f(x)=2(x+3)^2-8$

b) The vertex of the given function

The vertex of the parabola having form,

$\begin{gathered} y=ax^2+bx+c \\ \text{Vertex}=\frac{-b}{2a} \\ f(x)=2x^2+12x+10 \\ \text{Vertex}=\frac{-b}{2a}=\frac{-12}{2(2)}=-\frac{12}{4}=-3 \\ \text{Put x=-3 in }f(x)=2x^2+12x+10 \\ f(x)=2(-3)^2+12(-3)+10=18-36+10=-8 \end{gathered}$

Vertex is ( -3,-8)

c) x and y-intercept is,

$\begin{gathered} Set\text{ y=0 that means f(x)=0} \\ f(x)=2x^2+12x+10 \\ 2x^2+12x+10=0 \\ 2(x^2+6x+5)=0 \\ x^2+6x+5=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a},a=1,b=6,c=5 \\ x=\frac{-12\pm\sqrt{12^2-4\cdot\:2\cdot\:10}}{2\cdot\:2} \\ x=\frac{-12\pm\: 8}{4} \\ x=-1,x=-5 \end{gathered}$

x- intercepts are (-1,0) and (-5,0).

For y-intercept , set x=0

$\begin{gathered} f(x)=2x^2+12x+10 \\ y=2x^2+12x+10 \\ y=2(0)+12(0)+10 \\ y=10 \end{gathered}$

y-intercept is (0,10)

d) the graph of the function is,

e) The domain and range of the function is,

$\begin{gathered} \text{For range of the function f(x)=ax}^2+bx+c\text{ with vertex (x,y)} \\ \text{If a}<0\text{ range is }f(x)\leq y \\ \text{If a>0, range is f(x)}\ge\text{y} \end{gathered}$

For the given function,

$\begin{gathered} f(x)=2x^2+12x+10\text{ with vertex (-3,-8)} \\ a=2>0,\text{ range is f(x)}\ge\text{-8} \\ \text{Domain is -}\inftyTherefore,[tex]\begin{gathered} \text{Domain of f is (-}\infty,\infty) \\ \text{Range of f is \lbrack-8,}\infty) \end{gathered}$