Question im struggling with...

[Mr West]

Hey guys,

im only year 10 but i need some help with one of my math questions, and i reckon you guys may be able to help.

In the equation y=ax²+bx+c, find the values of a, b and c when the x-intercepts are (-4,0) and (1,0), and the y-intercept is (0,12).

Thanks

P.S wasnt sure which thread to put this in.

[Sine]

Hey guys,

im only year 10 but i need some help with one of my math questions, and i reckon you guys may be able to help.

In the equation y=ax²+bx+c, find the values of a, b and c when the x-intercepts are (-4,0) and (1,0), and the y-intercept is (0,12).

Thanks

P.S wasnt sure which thread to put this in.

[Mr West]

Thankyou so much! you are a legend!

[GMT. -_-]

Oh rip, Sine beat me

[Sine]

Thankyou so much! you are a legend!

there are also other ways to do this question
e.g. Subbing in each point that is give that the graph goes through (in this case the x-ints and y-ints) then create 3 simulateneous equation to solve for a,b,c.

[Mr West]

Sorry one more thing,

would you be able to show me how you made y=ax²+bx+c into y=d(x−e)(x−f)

[MightyBeh]

Here's the simultaneous equation method sine mentioned, sorry for the mess

Edit for this:

Sorry one more thing,

would you be able to show me how you made y=ax²+bx+c into y=d(x−e)(x−f)

This is the intercept/facotrised form, it comes from the null factor law that pretty much just means that the y value will be 0 when there's an x-intercept (makes sense). Putting that in to practice means that the term (x-e) will equal 0 at an x-intercept. Where sine subbed in (-4,0) for example, we have \(0 = (-4 -e) \), which implies \(e=-4\), and the same thing with (x-f). the d at the front is to make sure all the other points match up, without having that the equation you get at the end wouldn't necessarily go through the third point at (0,12)