Question

I need help with Graphing Quadratic Functions - *MEDAL*!
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Just show me how and answer a few.
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1.) y = x(2nd) -6x + 4



2.) y = 1/2x(2nd) + 6


3.) y = -x(2nd) + 4x - 4



4.) y + 4 = 3x(2nd)


-Just incase the (2nd) means (Power to the 2nd)-

Answer 1

hello?

Answer 2

So you need each of them answered?

Answer 3

I just need the steps but if you can do one and show me how it would be awesome.

Answer 4

not trying to make you do all the work :P

Answer 5

Mkay u mean y = x\(^{2}\)-6x + 4

Answer 6

yes

Answer 7

Okay that makes more sence, so u need this graphed?

Answer 8

Yup

Answer 9

https://www.desmos.com/calculator

Answer 10

Follow beccas steps lol plug them in and it will hsot u where it's graphed ;)

Answer 11

its not just that. I have to find the vertex and etc. until I find the point

Answer 12

Formula for finding the vertex:
\(\large x = \frac{-(b)}{2a}\)

For \(y = x^2 -6x + 4\), you know that the b term is -6 and the a term is 1. Just substitute those terms back to the formula.

\(\large x = \frac{-(-6)}{2(1)} = \frac{6}{2} = 3\)

Knowing that 3 is the x-coordinate for the vertex of the quadratic, we need to know what the y-coordinate is going to be. So now you need to plug it back into the equation.

\(y = (3)^2 - 6(3) + 4\)
\(y = 9 - 6(3) + 4\)
\(y = 9 - 18 + 4\)
\(y = -9 + 4\)
\(\color{green}{y = -5}\)

The vertex is (3, -5).

Answer 13

yeah I know that. I already have that on my paper, but there were a few more steps that I didn't really get to in class

Answer 14

Since the a value is positive, we can assume that the parabola will be facing upwards. Therefore, the vertex (3, -5) is the minimum point of the parabola. Now, start to plug some numbers in, but I suggest you to do it in this way: