find y-int, axis, vertex, etc; use these to graph x^2+8x+3

[cole92]

Ok, I've looked over my notes from class but still can't figure out what I'm doing wrong. I'll start a problem, check to see if it matches the answer in the back of my book to find out that I was only half right.

Here is the question (there are 6 problems I have to do but if I can get help on one, I can do the rest).

"Complete parts a-c for eac quadratic function.
a. Find the y-intercept, the equation of the axis of symmetry, and the x-coordinate of the vertex.
b. Make a table of values that includes the vertex.
c. Use this info to graph the function." ( I can do the graph if i can get help on parts a/b)

Here is one of the first functions:

f(x) = x^2 + 8x + 3

I get most of it done, but half of it is wrong...I'm not sure where I am making my error.

Any help would be greatly appreciated. Thanks in advance!

 

[kasie-tutor]

Re: help on quadratics please

Dear cole92,

You say that you tried this and only got it 1/2 right? We'd love to see your work (it's much more fun to show you where you went wrong than guess where your stuck).

Find y-intercept

The y-intercept is the coordinate where the x-value is zero. You have an equation: \(\displaystyle $f(x)=x^2+8x+3$\)
that will tell you any y-value if you input an x-value. So, TO FIND THE Y-INTERCEPT, plug in x=0 into the function. The y-intercept is \(\displaystyle $(0, f(0))$\). That is, (0, resulting y).

Find axis of symmetry

With your equation is in standard form (\(\displaystyle $ax^2+bx+c$\) form), the equation for the axis of symmetry is:
x = -B/(2A). So, what is B? What is A? Don't forget that the answer to the equation for the axis of symmetry is an equation. (Don't forget the x = in your answer).

Find the x-coordinate of the vertex

The x-coordinate of the vertex is given by -B/(2A). (The vertex occurs at the axis of symmetry.)

 

[cole92]

Re: help on quadratics please

ok...well I'm still not sure if I completely understand so I'll show my steps on what I would do:

f(x) = x[sup:3ld3o3dq]2[/sup:3ld3o3dq] + 8x + 3

h = -b/2a
h = (-8)/2(1)
h = -8/2
h = -4

so my vertex as of now is (-4, ?)

now I have to make my table:

X Y
-6 ?
-4 ?
-2 ?


Now what I did next is where I start to go wrong:
I think I am supposed to take -6 and -2 (my chosen points) and plug them into the function:

f(-6)= (-6)[sup:3ld3o3dq]2[/sup:3ld3o3dq] + 8(-6) + 3
= 36 - 48 + 3
f(-6) = -9

f(-4) = (-4)[sup:3ld3o3dq]2[/sup:3ld3o3dq] + 8(-4) + 3
= 16 - 32 + 3
= -13

f(-2) = (-2)[sup:3ld3o3dq]2[/sup:3ld3o3dq] + 8(-2) + 3
= 4 - 16 + 3
= -9


so my chart would be:
X Y
-6 -9
-4 -13
-2 -9

this makes my vertex (-4, -13)....with an axis of symmetry equation being x=-4?
and then of course i graph?

is this correct?

 

[kasie-tutor]

Re: help on quadratics please

so my chart would be:
X Y
-6 -9
-4 -13
-2 -9

Yes. I usually tell people to plot 5 points instead of three, but this is right.

this makes my vertex (-4, -13)....with an axis of symmetry equation being x=-4? Yes.
and then of course i graph?

Yes. Graph the equation AND the axis of symmetry (use a dotted line for the axis).

is this correct? Correct. And did you get (0, 3) for the y-intercept?