Quadratic functions

[HNO]

1. The vertex of a parabola is located at (-5, 6) and the y intercept is 231.

I am instructed to always use vertex form unless it asks for standard form.

I know that this is the base

(x+5) ^ squared

I just am unaware of how to use 231!

Thanks in advance (=

 

[mmm4444bot]

I am instructed to always use vertex form unless it asks for standard form.

Do you know what vertex form is? (Check out the link.)

Next, substitute your known values for y, x, h, and k into the vertex form, and then solve for a. :cool:

 

[HallsofIvy]

\(\displaystyle y= a(x- x_0)^2+ y_0\)
If you set \(\displaystyle x= x_0\), \(\displaystyle y= a(0)+ y_0= y_0\) no matter what a is. Further, if x is any number except \(\displaystyle x_0\), \(\displaystyle (x- x_0)^2\) is always positive so \(\displaystyle a(x- x_0)^2\) is always positive (if a is positive) or always negative (if a is negative). That is enough to say that \(\displaystyle (x_0, y_0)\) is the vertex. You can now use the fact that when x= 0 y= 231 to find a.

 

[HNO]

quadratic functions

Do you know what vertex form is? (Check out the link.)

Next, substitute your known values for y, x, h, and k into the vertex form, and then solve for a. :cool:

Yep I do know what vertex form is!

I have a(x+5)squared + 6 the answers for a is either -9 or 9…. but I don't exactly know where to go from where I am?

 

[math1]

y= a(x+5)² +6

substitute, (0, 231) in the equation, as y-intercept is 231.

we got, a=9

the equation is

y = 9(x+5)² + 6

 

[mmm4444bot]

Yep I do know what vertex form is!

I have a(x+5)squared + 6

That's only the right-hand side of the vertex form of a quadratic equation.

You need to write the complete equation.

:idea: If you read something here that you do not understand, then you should ask a specific question about it. (You skipped over the substitution of x and y.)

Also, why are you still typing the word "squared"?

It seems like you're not paying enough attention.