Another parabola turning point

Question

I have come across another one I am unable to do- I believe I am able to solve it but it doesn't work with the y-intercept I got. Please tell me how to solve for the turning point: $$y=-x^2-2x+15$$ For the y-intercept I got (0,15) and it is a maximum parabola. Thank you!

Answer 1

Your quadratic is $y=-x^2-2x+15$, from the form $ax^2+bx+c$, we see that $$a=-1\\b=-2\\c=15\tag{1}$$ Therefore, the $x$-value of the vertex is $$-\frac {b}{2a}=-\frac {-2}{-2}=-1$$ Plugging that back into the quadratic gives us the $y$ value: $$f(-1)=-(-1)^2-2(-1)+15=-1+2+15=16$$ Thus, the vertex is $$V=(-1,16)$$

---

Note that if you wanted to use Vertex form, the form $y=a(x-h)^2+k$, then you would first get the $b$ term, divide it by $2$, square it (completing the square) and combine all the like terms.

More specifically, we have $$f(x)=-x^2-2x+15\implies -(x^2+2x)+15$$$$\implies -(x^2+2x+1-1)+15\implies -(x+1)^2+16$$ The vertex is $(h,k)$ so we have $$V=(-1,16)$$ once more.