Question

Find a function whose graph is a parabola with vertex
(1, −5)
and that passes through the point
(3, 15).

Answer 1

This parabola is facing up you can tell by the fact that the vextex is below a point

Answer 2

I know how to do it I think.

Answer 3

oh ok

Answer 4

Gotta find the A

Answer 5

find it

Answer 6

y=ax^2+bx+c
y'=2ax+b
0=2ax+b
x=-b/2a
now u know that x=1, as that is the x value of the vertex so
1=-b/2a
-b=2a
now you also know that -5 is the value of y there, so
5=a(-b/2a)^2+b(-b/2a)+c <--- sub in -b=2a in there and finally solve for c with the final initial value

Answer 7

\(\bf y={\color{brown}{ a}}(x-h)^2+k\qquad vertex\ ({\color{blue}{ 3,15}})\to y={\color{brown}{ a}}(x-{\color{blue}{ 3}})^2{\color{blue}{ +5}}
\\ \quad \\
\textit{what's }"a"?\qquad \textit{passes through (3,15)}\qquad thus
\\ \quad \\
{\color{olive}{ 15}}={\color{brown}{ a}}({\color{olive}{ 3}}-{\color{blue}{ 3}})^2{\color{blue}{ +5}}\)