It is given to you that 4x+10y= -26 is an equation of the tangent line to the graph of y=g(x) at x=16.

It follows that y=__________

is the equation of the tangent line to y=g(x^2) at x=4

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You need to remember that the equation of the tangent line to the graph at a certain point is:

`y - y_0 = f'(x_0)(x - x_0)`

The given point is `x_0 = 16 ` and the given tangent line is `4x + 10y = -26.` You need to write the slope intercept form of the given equation of the tangent line, hence you should isolate y to the left side such that:

`10y = -4x - 26`

You need to divide by 10 both sides such that:

`y = -(2/5)x - 13/5`

You need to substitute `16` for `x_0` in equation `y - y_0 = f'(x_0)(x - x_0)` such that:

`y - g(x_0^2) = g'(x_0^2)(x - 16)`

`y = g'(x_0^2)*x - 16*g'(x_0^2) + g(x_0^2)`

You set equations `y = g'(x_0^2)*x - 16*g'(x_0^2) + g(x_0^2) ` and `y = -(2/5)x - 13/5` equal such that:

`g'(x_0^2)*x - 16*g'(x_0^2) + g(x_0^2) = -(2/5)x - 13/5`

`g'(x_0^2) = g'(16) = -2/5`

`- 16*g'(x_0^2) + g(x_0^2) = -13/5`

`-16*(-2/5) + g(x_0^2) = -13/5`

`g(x_0^2) = -32/5 - 13/5`

`g(x_0^2) = -45/5 =gt g(x_0^2) = -9`

**Hence, evaluating the equation of the tangent line to `y=g(x^2) at x=4` yields `y +9 = (-2/5)(x - 16).` **

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