We have been given graph of a downward opening parabola with vertex at point [tex](6.5,42.25)[/tex]. We are asked to write equation of the parabola in standard form.
We know that equation of parabola in standard form is [tex]f(x)=ax^2+bx+c[/tex].
We will write our equation in vertex form and then convert it into standard form.
Vertex for of parabola is [tex]y=a(x-h)^2+k[/tex], 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:
[tex]0=-a(0-6.5)^2+42.25[/tex]
[tex]0=-a(42.25)+42.25[/tex]
[tex]-42.25=-a(42.25)+42.25-42.25[/tex]
[tex]-42.25=-42.25a[/tex]
[tex]a=1[/tex] Divide both sides by [tex]-42.25[/tex]
So our equation in vertex form would be [tex]f(x)=-(x-6.5)^2+42.25[/tex].
Let us convert it in standard from.
[tex]f(x)=-(x^2-13x+42.25)+42.25[/tex]
[tex]f(x)=-x^2+13x-42.25+42.25[/tex]
[tex]f(x)=-x^2+13x[/tex]
Therefore, the equation of function is standard form would be [tex]f(x)=-x^2+13x[/tex].
