### Home > INT3 > Chapter 7 > Lesson 7.2.1 > Problem7-61

7-61.

For each function, determine the $x$‑ and $y$‑intercepts and the locator point ($h, k$). If you have not already done so, write the equations in graphing form.

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

$y$-intercept: $y = 7 + 2\left(0\right)^{2} + 4\left(0\right) − 5$

$x$-intercept: $0 = 7 + 2x^{2} + 4x − 5$

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

$y$-intercept: $\left(0, 2\right)$
$x$-intercept: $\left(−1, 0\right)$
vertex: $\left(−1, 0\right)$
$y = 2\left(x + 1\right)^{2}$

1. $x^{2} = 2x + x\left(2x - 4\right) + y$

Start by determining 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.