How do I put y=-5x^2-40x-6 in intercept form
can it be divided into 2 equations? Because that is the only possible way that this equation is making sense to me
I think so
yeah this question seems easy but is actually not
Hi! what is "intercept form"?
y=mx+b she/he meant slope intercept form
lol no
it is not a line, so it will not look like \(y=mx+b\)
lol yeah well i came here to answer because it appeared easy and now I want to know how you solve so yeah... if you know how then explain please :)
maybe it means "factor"
but this one does not factor with integers, so who knows?
yeah I tried to divide it into (x-a)(x-b)=c but it didnt gave an answer either
Slope intercept form only applies to straight lines.\[y=-5x^2-40x-6\]is not a straight line, and so it can't be represented using slope intercept form.
You can factor\[y=-5x^2-40x-6\]to become:\[y=-(5x^2+40x+6)=-(5x+40x+6)=-[(5x+0)(x+8)+6]=-(5x+0)(x+8)-6\]
@mathmate
For a quadratic, intercept form means f(x)=a(x-p)(x-p) where p, q are the roots of the function. |dw:1420919030082:dw| As @misty1212 pointed out, this one does not have rational factors, so it is less obvious.
To work out this problem, we need to solve the quadratic and find the two irrational roots p,q. Using the quadratic formula for f(x)=\(-5x^2-40x-6\) a=-5, b=-40 c=-6 x=\(\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) =\(\dfrac{40\pm\sqrt{40^2-4(-5)(-6)}}{2(-5)}\) =\(\dfrac{20\pm\sqrt{370}}{5}\) Putting the roots into the intercept form gives f(x)=\(-\dfrac{1}{5}(5x+(\sqrt{370}+20))(5x-(\sqrt{370}-20))\) which is the final intercept form. You should always expand the form to confirm that it is indeed f(x)=\(-5x^2-40x-6\) I will leave that to you as part of the exercise. I have exceptionally provided the final answer, partly because the arithmetic is tedious, and partly because I suspect there is a typo in the question. So whenever we come to a result like this, it is always a good idea to double check the question, in exams as well as in exercises.
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