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# Questions

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Find a+b+c if the graph of the equation $$y=ax^2+bx+c$$  is a parabola with vertex (5,3) , vertical axis of symmetry, and contains the point (2,0).

mathtoo  Mar 12, 2017
#1
+87334
+5

Find a+b+c if the graph of the equation   is a parabola with vertex (5,3) , vertical axis of symmetry, and contains the point (2,0)

We have that

y = a(x - h)^2 + k  and we know that (h, k)  = (5,3)  so

y = a(x - 5)^2 + 3       and since the point (2,0) is on the graph, we have that

0 = a (2 - 5)^2 + 3

0 = a(-3)^2 + 3

0 = 9a + 3

-3 = 9a        divide through by 3

-1 = 3a

-1/3  = a

So...our function is

y = (-1/3) (x - 5)^2 + 3     simplify

y = (-1/3) (x^2 - 10k + 25) + 3

y = (-1/3)x^2 + (10/3)x - 25/3 + 3

y = (-1/3)x^2 + (10/3)x - 25/3 + 9/3

y = (-1/3)x^2 + (10/3)x - 16/3

a = (-1/3)  b = (10/3)  and c = (-16/3)    so

a + b + c  =   [ -1 + 10 - 16] / 3  =  -7/3

Here's the graph : https://www.desmos.com/calculator/chosahxdr6

CPhill  Mar 12, 2017
#2
+7348
0

Find a+b+c if the graph of the equation is a parabola with vertex (5,3) , vertical axis of symmetry, and contains the point (2,0).

$$f(x)=y=ax^2+bx+c$$

$$c=3$$

$$x_1=2$$

$$x_2=8$$

$$a \ is \ negative$$

$$2 = {-b + \sqrt{b^2-12a} \over 2a}$$                $$8 = {-b- \sqrt{b^2-12a} \over 2a}$$

$$16a = {-b - \sqrt{b^2-12a} }$$

.         plus

$$4a = {-b + \sqrt{b^2-12a} }$$

$$20a=-2b$$

b=-10a

$$4a = {10a - \sqrt{100a^2-12a} }$$

$$-6a= - \sqrt{100a^2-12a}$$

$$36a^2=100a^2-12a$$

$$64a^2=12a$$

$$a=\frac{3}{16}$$  Error. CPhill was too fast.

asinus  Mar 12, 2017