Label the vertex and y-intercept (2 questions):
y= x^2 + 4x + 8

and

y= (x+4)^2 - 9

Solve (and graph) the inequalities (2 questions)
x^2 - 13x + 30 is < or = 0

and

x^2 + 2x - 35 > 0

Yes, I've gotten out my pencil and tried to do this. Yes, I'm asking for the worked out problems so that I can finish the rest of them by myself. Thanks!

I forget how to solve for the vertex but you can simply Google that. As for the y-intercepts, just plug in 0 for x.
And for the inequalities, you can just use the values it gives you and use them to plot the graphs.

I appreciate your input but I have 'simply Google'd these things, and it's incredibly difficult for me to learn math without seeing it first hand, ever since I rejected it in the fourth grade.

im an idiot and i dont have a clue what all that means

Don't forget dotted line for non-inclusive inequalities and solid lines for inclusive inequalities, as well as shading the proper side of the line to show where the answers lie (at least that's what I had to do back in the day)
Then why even bother posting?

Being lost with the first response, I haven't even found the way to yours. ; ;

i dont have a clue about that either, sorry

1+1=4. Self taught baby.

NO ITS ELEVENTY

Graph the first inequality like its a regular equation. For the first one, use all x values less than or equal to 0. Then the second inequality use all x values greater than 0. Then follow defrons advice.

The x-value of the vertex is half-way between the two x-intercepts. You then plug that in to get the y-value. Just like you get the y-intercept by setting x=0, you get the x-intercepts (two, since it's squared) by setting y=0

Both equations represent parabolas. Note that the vertex is the point at which the curve meets its axis. So, if you shift the origin to the vertex, the co-ordinates of the vertex become (0,0).

Now, y = x²+4x+8 needs to be expressed in terms of a general parabolic equation. i.e x²=(constant)y

=> y = (x²+4x+4)+4
=> (y-4) = (x+2)²

Now, I'll shift the origin to (-2,4). So that X = 0, Y = 0, represent the vertex, where X = x+2; Y= y-4,

X = 0, Y = 0.

=> x+2 =0, y-4 = 0
=> x = -2, y = 4 ....(those are the co-ordinates of the vertex)

The y-intercept is point at which the curve cuts the y-axis, simply put x=0 in the original equation to get the intercept.


=> x² - 13x + 30 <=0

You'll need to determine the factors of the equation. In this case,

=> (x-3)(x-10) <=0

Now, plot those numbers along a number line,

<-----(+)------(3)-------(-)--------(10)-----(+)----->

If you plot this quadratic, you'll notice that between the roots i.e. the interval (3, 10), the quadratic is -ve but for x = {3, 10}, the quadratic is 0. Also, note that for the interval x<3 and x>10, the quadratic is +ve (hence the signs on the above number line.)

So, the solution set would be,

=> x = [3, 10]

Edit: Forgot the inequality.

Thanks. I tried to comprehend what you said as much as possible.

Closed per OP request.