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The y coordinate of the vertex of f (x)=-x^2+8x+7 is f(4)

The Y Coordinate Of The Vertex Of F Xx28x7 Is F4 class=

Sagot :

limint602

Hello there.

To answer this question, we need to remember some properties about the vertex of quadratic functions.

Let [tex]f(x)=ax^2+bx+c,~a\neq0[/tex]. Its vertex can be found on the coordinates [tex](x_v,~y_v)[/tex] such that [tex]x_v=-\dfrac{b}{2a}[/tex] and [tex]y_v=-\dfrac{b^2}{4a}+c[/tex], in which [tex]y_v=f(x_v)[/tex].

Using the coefficients given by the question, we get that:

[tex]x_v=-\dfrac{8}{2\cdot(-1)}\\\\\\ x_v=-\dfrac{8}{-2}\\\\\\ x_v=4[/tex]

Thus, we have:

[tex]y=f(x_v)=f(4)[/tex]

So the statement is true, because the [tex]x[/tex] coordinate of the vertex of the function is equal to [tex]4.~~\checkmark[/tex]

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