1. ## Graph!

Help....

1. Graph y=-x^2+2x+1
Coordinates of all the intercepts:

Coordinates of the vertex

Can someone work through these and help...-thanks!

2. Originally Posted by mathhelplz
Help....

1. Graph y=-x^2+2x+1
Coordinates of all the intercepts:

Coordinates of the vertex

Can someone work through these and help...-thanks!
To find the intercepts of any function set the opposite variable equal to zero and solve the resulting equation

i.e x-intercept set y=0 solve for x to complete the ordered pair.

y-intercept set x=0 and solve for y.

There are a few different ways to find the intercept, what method are you using?

Good luck.

3. Originally Posted by mathhelplz
Help....

1. Graph y=-x^2+2x+1
Coordinates of all the intercepts:

Coordinates of the vertex

Can someone work through these and help...-thanks!
this is a downward opening parabola.

to find the x-intercepts, set y = 0 and solve for x. to find the y-intercept, set x = 0 and solve for y

you can find the vertex either by completing the square (if a parabola is written in the form $y = a(x - h)^2 + k$, then the vertex is $(h,k)$), or using the formula that for the vertex, $x = \frac {-b}{2a}$. thus the vertex occurs at $\left(\frac {-b}{2a}, f \left( \frac {-b}{2a}\right) \right)$ where $f(x)$ is the parabola