Find the values of k for which the x-axis is a tangent to the curve y=3x^2 - 8x + 5 - k. For each value of k, find the coordinates of the point of tangency.

Question

This question is asking for the range of values for k using the discriminant, while x=0 since it’s the x-axis. Thanks for helping!! 

Answer 1

The graph is tangent to the x-axis when y = 0.

0 = 3x² - 8x + 5 - k  is a parabola. The vertex is the point of tangency

0 = 3(x² - 8x/3 + (-4/3)²) + 5 - k - 16/3

0 = 3(x - 4/3)² - k - 1/3

 

-k - 13 = 0

 

Can you solve for k and answer?

Answer 2

x axis will be a tangent if the tangent point is the vertex. In that case, the equation 3x2-8x+5-k=0 would have only one  real solution, meaning

D=b²-4ac=(-8)²-12(5-k)=0. From here it follows that k=-1/3.

 

64-60+12k=0

12k=-4

k=-1/3