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The vertex is $(-4, 2)$ and the y-int is $(-3,-1)$. How would I solve this, or find out what "$a$" is so I can write the equation and graph it?

Daniel F. Thornton
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user133126
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  • The y intercept should lie on the $y$ axis. $(-3,-1)$ doesn't, so it isn't the $y$ intercept. – Batman Mar 04 '14 at 22:46

2 Answers2

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You have the equation $$ y = a(x-p)^2 + q $$ Note that the vertex for this quadratic is at $(p, q)$, so we know that $p = -4$ and $q = 2$. This gives us: $$ y = a(x+4)^2 + 2 $$ Now, plugging in our point $(-3, -1)$, we get: $$ -1 = a(-3+4)^2 + 2 \Rightarrow a = -3 $$ Therefore our equation is: $$ y = -3(x+4)^2 + 2 $$

kmbrgandhi
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  • Thanks for the help, the p and q is something i do understand.

    the question that i have to answer is "Set the quadratic function of the graph going through the point (-3, -1) and a peak / valley is at (-4,2)."

    the vertex is (-4,2) then right? and the y-intercept (-3,-1). When i do do the math answer like you gave me it says not good. Sorry for my english maybe i didn't asked my question right.

     – user133126 Mar 04 '14 at 22:56
  • I'm not exactly sure what you mean. Maybe the y intercept is supposed to be $(-3, 0)$? – kmbrgandhi Mar 04 '14 at 23:11
  • y-intercepts are of the form (0,a). x-intercepts are of the form (a,0). – Batman Mar 05 '14 at 00:09
  • Woops, random typo there: I meant $(0, -1)$. – kmbrgandhi Mar 05 '14 at 00:17
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The vertex form of a parabola tells you that $(p,q)$ is the vertex of the parabola (so $p=-4, q=2$). Now, plug in the other point, and solve for $a$.

Batman
  • 19,390
  • Thanks for the help, the p and q is something i do understand. the question that i have to answer is "Set the quadratic function of the graph going through the point (-3, -1) and a peak / valley is at (-4,2)." the vertex is (-4,2) then right? and the y-intercept (-3,-1). When i do do the math answer like you gave me it says not good. Sorry for my english maybe i didn't asked my question right. – user133126 Mar 04 '14 at 23:08