# Finding the turning point of a parabola

I know how to find the turning point of a parabola in most equations but I'm not sure how to solve it in this form, if anyone can help me please do! $$y=x^2+4x-5$$ Thanks

• Hint: Complete the square. – Jacky Chong Oct 9 '16 at 2:34
• The vertex of the quadratic of form $ax^2+bx+c=0$ Can be found by letting $x$ equal $-\frac {b}{2a}$ – Frank Oct 9 '16 at 2:47
• I'm curious what aspect of this problem is giving you problems! – Hurkyl Oct 9 '16 at 7:24

$$x^2 + 4x - 5 = (x^2 + 4x + 4) - 4 - 5 = (x + 2)^2 - 9.$$
Therefore, the vertex is at $(-2, \, -9)$.
The vertex can be found by plugging $x$ with $-\frac {b}{2a}$ give the form $ax^2+bx+c=0$.
So with your example $x^2+4x-5=0$, we have $$a=1\\b=4\\c=-5\tag{1}$$ So $-\frac {b}{2a}=-\frac {4}{2}=-2$. Plugging that into the quadratic gives $$f(-2)=4-8-5=-9\tag{2}$$ Therefore, the vertex is $(-2,-9)$.