Question

Any one know how to Graph Quadratic Equation?
Question are in the pictures please help me <3

Answer 1

@directrix

Answer 2

The x-squared term is positive so the parabola opens upward.

Answer 3

A?

Answer 4

You want the parabola that:
opens upward
crosses the x-axis at 4 and -2
and
has a vertex at (1, -9).

Do any of the options satisfy all three of those conditions? @KathyMae95

Answer 5

i think so a?

Answer 6

You don't need help completing the square then?

Answer 7

Not Option a because it does not cross the x-axis at all.

Which option crosses the x-axis at 4 and -2?

Answer 8

i though they where all down the rest

Answer 9

i say d i think im wrong so its c?

Answer 10

How about this? Does it meet all the criteria?

Answer 11

no?

Answer 12

in order to graph a quadratic function, you'll have to first study the domain, well... Any quadratic equation has no point of indetermination, but presenting a domain is indeed more "complete" in that sense.
Now, we write the domain as:
\[d(f)= \mathbb{R} \]
explaining that the domain is indeed all the real numbers.
I believe you don't have to use calculus for this so I'll stick with algebra.
Let's first look at the "roots" of the function, meaning the value of "x" when the function intersects the x-axis.
meaning that we want the image, to be zero:
\[x^2 -2x -8=0\]
Well, you know how to solve these, but it's with the general formula:
\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
\[x=\frac{ 2 \pm \sqrt{2^{2}-4(1)(-8)} }{ 2(1) }\]
\[x= \frac{ 2 \pm \sqrt{36} }{ 2 }\]
\[x_1 =4\]
\[x_2 =-2\]
We could find two solutions, so therefore, the function actually does cut the x-axis, So option A is not correct.
Let's now study the sign:

So, the parabola opens upward, so therefore, option "B", is not the answer.
Finally, the y- interception, calculated by making all the "x's" zero:
\[f(0)=0^2 -2(0) -8\]
\[f(0)=-8\]
So we discard the option "D".
Leaving us only with option "C".

Answer 13

so c it correct?

Answer 14

Yes.

Answer 15

yup.

Answer 16

ok thank u