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Mathematics 16 Online
OpenStudy (anonymous):

The minimum point is (-1,-2) and the y intercept is 4. find the equation of the quadratic function.

OpenStudy (anonymous):

is it y = 6(x+1)^2 -2 ?

OpenStudy (anonymous):

You are given the vertex as (-1, -2). Now let's write a quadratic equation is vertex form:\[\bf y=a(x-h)^2+k\]Where \(\bf (h,k)\) is the vertex. We are given that \(\bf (h,k)=(-1,-2)\) so let's plug that in:\[\bf \implies y= a(x-(-1))^2-2 =a(x+1)^2-2\]But we are also given that the y-intercept is 4 and recall that the y-intercept occurs when \(\bf x=0\) so by setting \(\bf x=0\) we get:\[\bf 4=a(0+1)^2-2 \implies 4=a-2 \implies a = 6\]Hence the equation becomes:\[\bf y=6(x+1)^2-2\]

OpenStudy (anonymous):

@kmeds16

OpenStudy (anonymous):

yey we have the same answer :D Haha thanks!

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