Write a quadratic function whose graph has a vertex of (3,2) and passes through the point (4,7). f(x)=

Answer:

y=5(x-3)^2+2

Step-by-step explanation:

You plug the point into f(x)=a(x-3)^2+2... that already has already been solved for the vertex that you want. Then you swap it out for the solution you have solved for.

The required function is [tex]f(x)=5(x-3)^2+2[/tex].

Important information:

The vertex form of a quadratic function is:

[tex]f(x)=a(x-h)^2+k[/tex]

Where, [tex]a[/tex] is a constant and [tex](h,k)[/tex] is vertex.

Substitute [tex]h=3,k=2[/tex].

[tex]f(x)=a(x-3)^2+2[/tex]         ...(i)

The graph passes through the point (4,7). Substitute [tex]x=4, f(x)=7[/tex] in (i).

[tex]7=a(4-3)^2+2[/tex]

[tex]7=a+2[/tex]

[tex]7-2=a[/tex]

[tex]5=a[/tex]

Substitute [tex]a=5[/tex] in (i).

[tex]f(x)=5(x-3)^2+2[/tex]

Therefore, the required function is [tex]f(x)=5(x-3)^2+2[/tex].

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