f(x)=xsquared + 6x -4 Find the vertex, the y-intercept, and the equation of the axis of symmetry. Then graph the function. :confused:

Ask the teacher. If you want to learn, learn by going to the teacher so he/she can take you through step by step instructions. This'll be more easier when you get another question like this one

First - the y-intercept is the point where x = 0. So simply substitute x=0 and see what you get for f(0).

The next thing is to restate the equation into a standard form that is like this:

y-b = (x-a)^2

The reason is that if you can find the values of 'a' and 'b', then the vertex is at (a,b). You need to emorize this basic form. So, starting with:

y = x^2 +6x -4

First - complete the square - take half the value of the coefficient of the x term, square it, and then add and subtract that value on the right hand side. This is a trick that is very useful to have in your arsenal - it comes up a lot! Here the coefficient for the x term is 6, so half of that squared is 9:

y= x^2 + 6x + 9 -9 - 4 = x^2 - 6x + 9 -13

Bring the 13 to the left hand side:

y +13 = x^2 +6x + 9

Now you can factor the right hand side:

y+13 = (x+3)^2

Now you can see why completing the square was a good idea - it gets you to the form you wanted of (x-a)^2.

Note that the basic eqation has a negative sign in front of both the 'a' and 'b' terms, so we need to restate this as:

y - (-13) = (x - (-3))^2

So now we have a = -3 and b = -13. This tells you the vertex is at the point (-3,-13).
Finally, the axis of symmetry goes through the vertex, and is parallel to the y axis. Can you take it from here?