Question

Determine the vertex of (f)=-×^2+50×-456

Answer 1

Answer:

$V_x = -\frac{50}{2*(-1)} =25$

And for the coordinate on y we can use the function like this:

$V_y = -(25)^2 +50*25 -456 =169$

Then the vertex would be $V= (25,169)$

Step-by-step explanation:

For this problem we have the following function given:

$f(x) = -x^2 +50x -456$

That represent a quadratic function and the general form is given by:

$f(x)= ax^2 +bx +c$

For this problem a =-1 , b= 50 , c=-456 and we can find the corrdinate of the vertex in x with this formula:

$V_x =-\frac{b}{2a}$

And replacing we got:

$V_x = -\frac{50}{2*(-1)} =25$

And for the coordinate on y we can use the function like this:

$V_y = -(25)^2 +50*25 -456 =169$

Then the vertex would be $V= (25,169)$