We have been given graph of a downward opening parabola with vertex at point $(6.5,42.25)$ . We are asked to write equation of the parabola in standard form.

We know that equation of parabola in standard form is $f(x)=ax^2+bx+c$ .

We will write our equation in vertex form and then convert it into standard form.

Vertex for of parabola is $y=a(x-h)^2+k$ , where point (h,k) represents vertex of parabola and a represents leading coefficient.

Since our parabola is downward opening so leading coefficient will be negative.

Upon substituting coordinates of vertex and point (0,0) in vertex form, we will get:

$0=-a(0-6.5)^2+42.25$

$0=-a(42.25)+42.25$

$-42.25=-a(42.25)+42.25-42.25$

$-42.25=-42.25a$

$a=1$ Divide both sides by $-42.25$

So our equation in vertex form would be $f(x)=-(x-6.5)^2+42.25$ .

Let us convert it in standard from.

$f(x)=-(x^2-13x+42.25)+42.25$

$f(x)=-x^2+13x-42.25+42.25$

$f(x)=-x^2+13x$

Therefore, the equation of function is standard form would be $f(x)=-x^2+13x$ .

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3 years ago