Quadratics

[kris.yarmak123]

The zeros of a quadratic function are 4 and -3. If the graph of the function has a y-intercept of 48, the quadratic must be.... (write a formula)
Plz help

 

[MarkFL]

Consider that the family of quadratics having the given zeros can be given as:

[MATH]f(x)=k(x+3)(x-4)[/MATH] where \(0\ne k\in\mathbb{R}\)

We are told \(f(0)=48\), so can you find the value of \(k\) from this?

 

[kris.yarmak123]

Consider that the family of quadratics having the given zeros can be given as:

[MATH]f(x)=k(x+3)(x-4)[/MATH] where \(0\ne k\in\mathbb{R}\)

We are told \(f(0)=48\), so can you find the value of \(k\) from this?

K = -4 right?

 

[kris.yarmak123]

I

Consider that the family of quadratics having the given zeros can be given as:

[MATH]f(x)=k(x+3)(x-4)[/MATH] where \(0\ne k\in\mathbb{R}\)

We are told \(f(0)=48\), so can you find the value of \(k\) from this?

I got it thanks! The answer is -4x^2+4x+48

 

[JeffM]

You can and should learn to check your own work, but we are happy to confirm.

[MATH]f(x) = -\ 4x^2 + 4x + 48 \implies[/MATH]
[MATH]f(0) = -\ 4(0)^2 + 4(0) + 48 = 0 + 0 + 48 = 48 \text { check,}[/MATH]
[MATH]f(4) = -\ 4(4)^2 + 4(4) + 48 = -\ 64 + 16 + 48 = -\ 64 + 64 = 0 \text { check,}[/MATH]
[MATH]f(-\ 3) = -\ 4(-\ 3)^2 + 4(-\ 3) + 48 = -\ 36 - 12 + 48 = -\ 48 + 48 = 0 \text { check.}[/MATH]
Good job.

 

[kris.yarmak123]

You can and should learn to check your own work, but we are happy to confirm.

[MATH]f(x) = -\ 4x^2 + 4x + 48 \implies[/MATH]
[MATH]f(0) = -\ 4(0)^2 + 4(0) + 48 = 0 + 0 + 48 = 48 \text { check,}[/MATH]
[MATH]f(4) = -\ 4(4)^2 + 4(4) + 48 = -\ 64 + 16 + 48 = -\ 64 + 64 = 0 \text { check,}[/MATH]
[MATH]f(-\ 3) = -\ 4(-\ 3)^2 + 4(-\ 3) + 48 = -\ 36 - 12 + 48 = -\ 48 + 48 = 0 \text { check.}[/MATH]
Good job.

Yeah, I know how to check, we just did it late in February and I kinda forgot it a little bit. Thank you very much!