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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

#1
+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
Sort:

#1
+17614
+3

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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