# `y = (x-2)(x^2 + 3x), (1,-4)` (a) Find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point,...

`y = (x-2)(x^2 + 3x), (1,-4)` (a) Find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results.

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**(a)** Take the derivative. Use the product rule.

`y'=(x-2)(2x+3)+(x^2+3x)(1)= (x-2)(2x+3)+(x^2+3x)`

Given the point `(1,-4)` , substitute `x=1` .

`y'(1)= (1-2)(2(1)+3)+(1^2+3(1)) = -1(5)+(4)=-1`

The slope at is -1.

Write the slope intercept form, substitute the point and the slope to find b.

`y=mx+b`

`-4=(-1)(1)+b`

`b=-3`

The equation of the tangent line is: `y=-x-3`

**(b) Graph: This is only the original function... The graph will not allow me to plot y=-x-3 to add on this graph, but this function should show the tangent line on the graph. I have attached another image to the answer.**

**(c) You can do this on your own. It's the dy/dx function.**