What is the asymptote of the graph of h(x)=4x−2?

There is no asymptote

Step-by-step explanation:

Asymptotes occur when you have a divide by 0 error/undefined. The graph of h(x) = 4x - 2 is a linear line, so no asymptotes are present.

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1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.