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).
