Graph: y=f(x)=-x2+3x. Give the coordinates of the vertex, the x-intercepts (if they exist), the y-intercept, the domain, range, max/min value, and the equation of the axis of symmetry.

vertex
x-int.
y-int.
domain
range
max/min
axis of symmetry

2. Originally Posted by Jluse7
Graph: y=f(x)=-x2+3x. Give the coordinates of the vertex, the x-intercepts (if they exist), the y-intercept, the domain, range, max/min value, and the equation of the axis of symmetry.

vertex
x-int.
y-int.
domain
range
max/min
axis of symmetry
Hi Jluse7,

You need to translate your functional notation into vertex form. The graph will be a parabola opening downward.

$f(x)=-x^2+3x$

$f(x)=-(x^2-3x)$

Complete the square:

$f(x)=-\left(x^2-3x{\color{red}+\frac{9}{4}}\right){\color{red}+\fr ac{9}{4}}$

$f(x)=\left(x-\frac{3}{2}\right)^2+\frac{9}{4}$

This tells you that the vertex of your parabola is at $\left(\frac{3}{2}, \frac{9}{4}\right)$

Set f(x) = 0 to find your x intercepts. Use your original function.

$f(x)=-x^2+3x=0$

$-x(x-3)=0$

$x=0 \ \ or \ \ x=3$

Notice, since x = 0, f(0) = 0, thus the y-intercept is 0.

The domain will be all real numbers. There are no restrictions on x.

The range will be $\left\{y|y \le \tfrac{9}{4}\right\}$

The maximum value will occur at the vertex where $y=\frac{9}{4}$.

There is no minimum.

The axis of symmetry is $x=\frac{3}{2}$. This is a vertical line which passes through the vertex.