How to find a tangent line that passes through two graphs?

Step 1: For a point (x,y)(x,y) on the graph of f(x)=x2f(x)=x2, find the slope of the line between (x,y)(x,y) and (3,8)(3,8).

Step 2: Compute the slope of the tangent line to f(x)=x2f(x)=x2 at the point (x,y)(x,y).

Step 3: Set these two slopes equal to each other and find candidate xx values.

Step 4: Check your answers.

Here is a problem to support that
hope thiz helps.

point on the graph is (x,x2)(x,x2). The slope from that point to (3,8)(3,8) is given by: x2−8x−3x2−8x−3. This has to be equal to the derivative at the point for it to be a tangent. So:

x2−8x−3=2xx2−8x−3=2x

x2−8=2x2−6xx2−8=2x2−6x

0=x2−6x+80=x2−6x+8

0=(x−2)(x−4)0=(x−2)(x−4)

So x=2x=2 or x=4x=4. So the points are (2,4)(2,4) and (4,16)(4,16).

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