Practice Test 4, Section 3, Number 11 (No Calc)

I love this question because the fastest way to go involves almost no math. You just have to know a little bit about the shapes of lines and parabolas.

First, think about the parabola. Its equation is y=(2x-3)(x+9). From that we know it’s a parabola that opens up (the x^2 term will be positive) and has x-intercepts at \dfrac{3}{2} and -9. You should figure out its y-intercept, too, by plugging in zero for x:

    \begin{align*}y&=(2(0)-3)(0+9)\\y&=(-3)(9)\\y&=-27\end{align*}

Do a very rough drawing of that on your paper (forgive my MS Paint skillz, but your drawing can be sloppier than mine and still be plenty good enough):

Now do a rough drawing of the line. To do that, put it in slope-intercept form:

    \begin{align*}x&=2y+5\\x-5&=2y\\y&=\dfrac{1}{2}x-\dfrac{5}{2}\end{align*}

The important detail there is that the y-intercept is -\dfrac{5}{2}, which is above the parabola’s y-intercept of -27, so you know the line will intersect the parabola twice. Like so: