# Math Help - Trying to solve for y here....

1. ## Trying to solve for y here....

I'm working on a calculus problem, but I'm stuck with something algebraically:

$x = y^2-7y$

I'm trying to solve for y so I can graph it. I started by factoring y:

$x = y(y-7)$

But that's as far as I can get. If I divide both sides by y, that still leaves me with x/y = y-7. The extra y on the right side of the equation is stumping me and I can't figure out how to get rid of it. Any ideas?

2. This is a quadratic equation: $y^2-7y-x=0.$ Depending on $x$, it may have zero, one or two solutions, which are given by the standard formula for the roots of a quadratic equation.

3. Originally Posted by bobbooey
I'm working on a calculus problem, but I'm stuck with something algebraically:

$x = y^2-7y$

I'm trying to solve for y so I can graph it. I started by factoring y:

$x = y(y-7)$

But that's as far as I can get. If I divide both sides by y, that still leaves me with x/y = y-7. The extra y on the right side of the equation is stumping me and I can't figure out how to get rid of it. Any ideas?
complete the square ...

$x + \frac{49}{4} = y^2 - 7y + \frac{49}{4}$

$x + \frac{49}{4} = \left(y - \frac{7}{2}\right)^2$

$x = \left(y - \frac{7}{2}\right)^2 - \frac{49}{4}$

vertex of the parabola is at $\left(-\frac{49}{4} , \frac{7}{2}\right)$ and the parabola opens up to the right.

4. My issue is that I need to find where the two lines intersect (In Quadrant II)

x = y^2-7y
x = 2y-y^2

After about 30 minutes of brute forcing it, I figured out that that they intersect at (-11.25,4.5) but does an easier method exist mathematically for determining that?

5. Originally Posted by bobbooey
My issue is that I need to find where the two lines intersect (In Quadrant II)

x = y^2-7y
x = 2y-y^2

After about 30 minutes of brute forcing it, I figured out that that they intersect at (-11.25,4.5) but does an easier method exist mathematically for determining that?

solve $y^2-7y= 2y-y^2$