How do you find the slope of the tangent line to the graph at given point and give an equation of the tangent line f(x)=32x at (1, 3/2)?

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Raina Blackburn · Accepted Answer

This function is a linear function. Its graph is the straight line&amp;nbsp;y=32x.The tangent line can be described as the limit, as a second point approaches(1,32), of the slope of secant line.For every point on the graph, the secant line is the same as the graph of the original function,&amp;nbsp;y=32x.Therefore, the tangent line is the same,&amp;nbsp;y=32x.If we want the details:The slope of the tangent line is given bylimh→0f(1+h)−f(1)h=limh→032(1+h)−32(1)h=limh→032hh=32The line through the point&amp;nbsp;(x1,y1)&amp;nbsp;with slop&amp;nbsp;m=32&amp;nbsp;can be writteny−y1=m(x−x1)So we havey−32=32(x−1)&amp;nbsp;Solving for y to get slope intercept form, we gety=32x

How do you find the slope of the

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