## anonymous one year ago the values of x for which the graphs of y=x+2,y^2=4x intersect are: a. -2 and 2 b. -2 c. 2 d. 0 e. none of these

1. zepdrix

$\large\rm y=x+2,\qquad\qquad y^2=4x$We have a couple ways to approach this... Hmm, let's try solving each equation for x.

2. zepdrix

In the first equation, subtract 2 gives us,$\large\rm y-2=\color{orangered}{x}$And in the second equation, dividing by 4 gives us,$\large\rm \frac{y^2}{4}=\color{orangered}{x}$They will intersect when these x's are the same, which means the equations are the same.$\large\rm y-2=\frac{y^2}{4}$

3. zepdrix

Multiply by 4,$\large\rm 4y-8=y^2$Move everything to one side,$\large\rm y^2-4y+8=0$

4. zepdrix

It looks like this is not going to factor nicely. Try using your quadratic formula:$\large\rm y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$Plug in the values, what does it tell you? :)

5. anonymous

so its b? @zepdrix

6. zepdrix

Hmm +_+ Well, we're trying to find our y-coordinate, and we can then use that to find the x-coordinate of intersection. What did you get for the y-value? Use the formula.

7. anonymous

-2 and 0 @zepdrix

8. jiteshmeghwal9

$$y=x+2$$ $$y^2=4x$$ From above eqn $$y^2=4(y-2)$$ $$y^2-4y+8=0$$ solve eqn either by quadratic formula or factorization.

9. jiteshmeghwal9

$y=\frac{4 \pm \sqrt{16-32}}{2}$and $x=y-2$

10. anonymous

so im right @jiteshmeghwal9

11. jiteshmeghwal9

$y=\frac{4 \pm 4 \iota}{2}=2\pm2 \iota$ $x=y-2=2\pm2 \iota -2$$x= \pm 2 \iota$