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The graph of the quadratic $y = ax^2 + bx + c$ has the following properties: (1) The maximum value of $y = ax^2 + bx + c$ is 5, which occurs at $x = 3$. (2) The graph passes through the point $(0,-13)$. If the graph passes through the point $(4,m)$, then what is the value of $m$?

michaelcai  Jul 21, 2017

Best Answer 

 #1
avatar+17614 
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If the x-value of the vertex (maximum value) is 3 and the y-value is 5, the vertex occurs at (3, 5).

 

An equation for a parabola is:  y - k  =  a(h - h)2       [ The vertex occurs at (h, k) ]

     --->                                      y - 5  =  a(x - 3)2

 

Since the graph passes through the point (0, -13), we have:

                                                -13 - 5  =  a(0 - 3)2

                                                     - 18  =  a(-3)2

                                                       -18  =  a·9

                                                          a  =  -2

So, the equation is:  y - 5  =  -2(x - 3)2

For the point (4, m),  x = 4  and  y = m   --->     m - 5  =  -2(4 - 3)2

Now, solve this for m.

geno3141  Jul 21, 2017
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1+0 Answers

 #1
avatar+17614 
+3
Best Answer

If the x-value of the vertex (maximum value) is 3 and the y-value is 5, the vertex occurs at (3, 5).

 

An equation for a parabola is:  y - k  =  a(h - h)2       [ The vertex occurs at (h, k) ]

     --->                                      y - 5  =  a(x - 3)2

 

Since the graph passes through the point (0, -13), we have:

                                                -13 - 5  =  a(0 - 3)2

                                                     - 18  =  a(-3)2

                                                       -18  =  a·9

                                                          a  =  -2

So, the equation is:  y - 5  =  -2(x - 3)2

For the point (4, m),  x = 4  and  y = m   --->     m - 5  =  -2(4 - 3)2

Now, solve this for m.

geno3141  Jul 21, 2017

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