FIND THE VERTEX THE LINE OF SYMMETRY THE MAX OR MIN VALUE OF THE FOLLOWING

F(X)= -X^2+10X+5

2 answers

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Finding the vertex of a parabola in standard form: f(x)=ax^2 + bx + c uses a formula to find the x-coordinate first:
x=-b/(2a)

x=-10/(2*-1)
x=-10/-2
x=5

To find the y-coordinate, substitute this newly found value of x into the original function.

f(5)=-(5)^2+10(5)+5 (careful with PEMDAS!)
f(5)=-25+50+5
f(5)=30

Therefore, your vertex is at: (5,30)

The Line(Axis) of Symmetry is easy from here. Parabolas given in the form y=x^2 will always be symmetric about a line that is vertical and MUST pass through the vertex. So the equation of the line of symmetry is: x = 5.

The Ymin or Ymax value depends on whether the coefficient of the x^2 term is positive or negative. If its positive, then the parabola opens UPWARD, thus giving you a MIN. If the coefficient is negative, then the parabola opens DOWNWARD, thus giving you a MAX.

So, you have a -x^2, so your parabola opens DOWN, giving you a Ymin. Its specific value is also connected to your vertex. It is equal to the y-coordinate of your vertex. So, its Ymin=30.

Hope this helps!