$\displaystyle f(x)=x^2+6x$
Thanks a lot for your answer, but I can't use the table of values anymore
Do you have to factor out the common x first to make it
x(x+6) and then graph it as a normal line, six units left?
or are you supposed to graph it as a quadratic function? :S
When you factor the x out, does the make whatever is in the brackets the only important thing now?
I am so confused with what I have to do if there is more than one x!
There are many methods. What method are you supposed to use.
Here is one...
We could first find the x-intercepts by setting y=0
$\displaystyle 0=x^2+6x=x(x+6) \implies x=0 or x=-6$
So the ordered pairs of the x intercepts are (0,0) and (-6,0)
Frome here we can find the vertex by using the fact that a parabola is symmetric about its vertex. That means the x coordinate of the vertex is at the mid point of the x intercepts. $\displaystyle x=\frac{0+(-6)}{2}=-3$
Now we can complete the order pair by plugging this back into the equation
$\displaystyle y=(-3)^2+6(-3)=9-18=-9$ so the vertex is at (-3,-9)
So if we plot these three points we can sketch the parabola.
Another methold we be to complete the square...
Do any of these sound familar...