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alibeatles · Oct 18, 2009, 08:31 AM

I can't remember how to tackle this type of question, I don't need the answer, just the steps.

Find the y-intercept of the parabola with a vertex of (-5,2) that passes through (-4,5)

Unknown008 · Oct 18, 2009, 10:32 AM

Well, I don't know how you would do it, but I would have used the completed square formula to solve this.

The completed square form of a general quadratic is [math]y = a(x + \frac{b}{2a})^2 - \frac{b^2}{4a} + c[\math]

The vertex is given by $( -\frac{b}{2a},\, - \frac{b^2}{4a} + c)$

That means, $-5 = -\frac{b}{2a}$ and $2 = - \frac{b^2}{4a} + c$

Then, using your given point, you have another equation; $(-4) = a((5) + \frac{b}{2a})^2 - \frac{b^2}{4a} + c$

You should be able to solve for a, b and c, then, set x to zero in your equation to have the y intercept.