Question 999447
You need to complete the square to put the equation in vertex form.
If you're unsure how to do this, search on "completing the square" on YouTube. 
{{{f(x)=x^2+2x-15}}}
{{{f(x)=(x^2+2x+1)-1-15}}}
{{{f(x)=(x+1)^2-16}}}
So the vertex is ({{{-1}}},{{{-16}}}).
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The vertex lies on the axis of symmetry {{{x=-1}}}.
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For the x-intercepts,
{{{0=(x+1)^2-16}}}
{{{(x+1)^2=16}}}
{{{x+1=0 +- 4}}}
{{{x=-1 +- 4}}}
{{{x=3}}} and {{{x=-5}}}
(3,0) and (-5,0)
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For the y-intercept,
{{{y=(0+1)^2-16}}}
{{{y=1-16}}}
{{{y=-15}}}
(0,-15)
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*[illustration v2.JPG].