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
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.
$\displaystyle f(x)=-x^2+3x$
$\displaystyle f(x)=-(x^2-3x)$
Complete the square:
$\displaystyle f(x)=-\left(x^2-3x{\color{red}+\frac{9}{4}}\right){\color{red}+\fr ac{9}{4}}$
$\displaystyle f(x)=\left(x-\frac{3}{2}\right)^2+\frac{9}{4}$
This tells you that the vertex of your parabola is at $\displaystyle \left(\frac{3}{2}, \frac{9}{4}\right)$
Set f(x) = 0 to find your x intercepts. Use your original function.
$\displaystyle f(x)=-x^2+3x=0$
$\displaystyle -x(x-3)=0$
$\displaystyle x=0 \ \ or \ \ x=3$
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 $\displaystyle \left\{y|y \le \tfrac{9}{4}\right\}$
The maximum value will occur at the vertex where $\displaystyle y=\frac{9}{4}$.
There is no minimum.
The axis of symmetry is $\displaystyle x=\frac{3}{2}$. This is a vertical line which passes through the vertex.