Answer:
[tex]y=-3(x-5)^2+4[/tex]
Step-by-step explanation:
Hi there!
Given the vertex of a parabola and a point, it's easiest to organize the equation in vertex form:
[tex]y=a(x-h)^2+k[/tex] where the vertex is located at [tex](h,k)[/tex] and a is a numerical value
1) Plug the vertex into the equation
[tex]y=a(x-h)^2+k[/tex]
Plug in the vertex (5,4)
[tex]y=a(x-5)^2+4[/tex]
2) Solve for a
[tex]y=a(x-5)^2+4[/tex]
Plug in the given point (3,-8) and solve for a
[tex]-8=a(3-5)^2+4\\-8=a(-2)^2+4\\-8=4a+4[/tex]
Subtract 4 from both sides
[tex]-8-4=4a+4-4\\-12=4a[/tex]
Divide both sides by 4
[tex]\frac{-12}{4} = \frac{4a}{4} \\-3=a[/tex]
Therefore, a=-3. Plug this back into [tex]y=a(x-5)^2+4[/tex]:
[tex]y=-3(x-5)^2+4[/tex]
I hope this helps!
