PQR is an isosceles triangle in which PQ = PR. M and N
1 answer:
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All we need is to put this form in the vertex form f(x) = (ax+b)^2 + c
So we have <span>f (x)= 3x^2+12x+11 ....
Let's complete the square (if you aware of it)
</span><span>
f(x)= 3x^2+12x+11 = 3(x^2+4x)+11 = 3(x^2+4x+4-4)+11
=</span><span> 3([x^2+4x+4]-4)+11 = 3[(x+2)^2-4]+11 =3</span><span>(x+2)^2 - 12 +11 = 3</span><span><span>(x+2)^2 -1
so our form would be:
Here is a parabola with vertex of (-2,-1) and with positive </span> slope (concave up)
</span>
So we have <span>f (x)= 3x^2+12x+11 ....
Let's complete the square (if you aware of it)
</span><span>
f(x)= 3x^2+12x+11 = 3(x^2+4x)+11 = 3(x^2+4x+4-4)+11
=</span><span> 3([x^2+4x+4]-4)+11 = 3[(x+2)^2-4]+11 =3</span><span>(x+2)^2 - 12 +11 = 3</span><span><span>(x+2)^2 -1
so our form would be:
Here is a parabola with vertex of (-2,-1) and with positive </span> slope (concave up)
</span>
I hope that helps!
