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Find the x-coordinate of the point on the graph of y=x^2 where the tangent line is parallel to the secant line that cuts the curve at x=-1 and x=2
Please how do you solve this?

Respuesta :

LammettHash
First find the secant line. The slope of the secant line through [tex](-1,1)[/tex] (when [tex]x=-1[/tex]) and [tex](2,4)[/tex] (when [tex]x=2[/tex]) is the average rate of change of [tex]y=x^2[/tex] over the interval [tex][-1,2][/tex]:

[tex]\text{slope}_{\text{secant}}=\dfrac{2^2-(-1)^2}{2-(-1)}=\dfrac33=1[/tex]

The tangent line to [tex]y=x^2[/tex] will have a slope determined by the derivative:

[tex]y=x^2\implies y'=2x[/tex]

Both the secant and tangent will have the same slope when [tex]2x=1[/tex], or when [tex]x=\dfrac12[/tex].

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