With this i have no idea how to don"t ignore this one helllpconsider the following graph of a quadratic function. write the euqation for the quadratic function in standard form. the graphs y-intercept is 7 and vertex is (-3,-2)

Write the vertex form of the equation and find the necessary coefficient to make it work.
.. y = a*(x +3)^2 -2
.. = ax^2 +6ax +9a -2

You require the y-intercept to be 7. So, for x=0, you have
.. 9a -2 = 7
.. 9a = 9
.. a = 1

The equation you seek is
.. y = x^2 +6x +7

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