Write the equations in graphing form, then state the vertex of the parabola or the center and radius of the circle.
1. f (x)= 3x^2+12x+11

All we need is to put this form in the vertex form f(x) = (ax+b)^2 + c

So we have f (x)= 3x^2+12x+11 ....

Let's complete the square (if you aware of it)

f(x)= 3x^2+12x+11 = 3(x^2+4x)+11 = 3(x^2+4x+4-4)+11
= 3([x^2+4x+4]-4)+11 = 3[(x+2)^2-4]+11 =3(x+2)^2 - 12 +11 = 3(x+2)^2 -1

so our form would be:



Here is a parabola with vertex of (-2,-1) and with positive  slope (concave up)

I hope that helps!

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12