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).
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
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.