Home > CCA2 > Chapter 2 > Lesson 2.2.3 > Problem2-128

2-128.

For each equation below, find the $x$- and $y$-intercepts and the locator point $(h,k)$, then write the equations in graphing form.

1. $y=7+2x^2+4x-5$

$y$-intercept: $y=7+2(0)^2+4(0)−5$

$x$-intercept: $0=7+2x^2+4x−5$

Since there is only one $x$-intercept, the vertex must be at the $x$-intercept.

$y$-intercept: $(0,2)$
$x$-intercept: $(−1,0)$
vertex: $(−1,0)$
$y=2(x+1)^2$

1. $x^2=2x+x(2x-4)+y$

Start by finding the $x$- and $y$-intercepts.

To find the $x$-coordinate of the vertex,
average the $x$-intercepts.

To find the $y$-coordinate of the vertex, substitute the
$x$-coordinate of the vertex into the given equation and solve for $y$.

Now write the equation in graphing form.