Chapter ISG, Problem 9.1.4P

To determine

To find intercept & vertex of given function $y=\frac{1}{2}{x}^2+2x$

Answer to Problem 9.1.4P

Intercept: x-intercept = $\pm 1$ and y-intercept = 2

Vertex: $\left(0,2\right)$

Explanation of Solution

Given Information: $y=\frac{1}{2}{x}^2+2x$

Formula Used: Graphing Utility

Calculation:

For such equations, a graphing utility can often be used to investigate possible solutions. When a graphing utility is used to solve an equation, usually approximate solutions are obtained.

  $y=\frac{1}{2}{x}^2+2x$

So, simplify

  $2y=x^2+4x$

Put the value of x in y, get:

x	0	-1	-2	-3	-4
y	0	-1.5	-2	-1.5	0

  

Hence,

Intercept: x-intercept = 0, -4 and y-intercept = 0

Vertex: $\left(-2,-2\right)$