Find the general expression for the slope of a line tangent to the curve of y=2x^2+4x at the point P(x,y) . Then find the slopes for x=-3 and x=0.5. Sketch the curve and the tangent lines. What is the general expression for the slope of a line tangent to the curve of the function y=2x^2+4 at the point P(x,y) ​?

Never mind I got it. mtan=6x+4. Sub in x values, for x=-2, take 6(-2)+4 = -8, and for x=1.5. 6(1.5)+4=13.

Complete Question

Find the general expression for the slope of a line tangent to the curve of y=2x^2+4x at the point P(x,y) . Then find the slopes for [tex] x = 3[/tex] and x=0.5. Sketch the curve and the tangent lines. What is the general expression for the slope of a line tangent to the curve of the function y=2x^2+4 at the point P(x,y) ​?

Answer:

The  generally expression for the slope of [tex]y = 2x^2 + 4x[/tex] is  [tex]y' = 4x +4[/tex]

The graph is shown on the first uploaded image

The  generally expression for the slope of [tex]y=2x^2+4[/tex] is   [tex]y' = 4x[/tex]

Step-by-step explanation:

From the question we are told that

  The  equation of the curve is [tex]y = 2x^2 + 4x[/tex]

First we differentiate the equation

So  

     [tex]y' = 4x +4[/tex]

Therefore the generally expression for the slope tangent to the curve [tex]y=2x^2+4x[/tex] is   [tex]y' = 4x +4[/tex]

The  next step is to substitute for x =  3 and  x =  0.5

So  for [tex]x_1 = 3 [/tex]

    [tex]y' = 4(3) +4[/tex]

     [tex]y' =m_1= 16[/tex]

And  for  [tex]x_2 = 0.5[/tex]

      [tex]y' = 4(0.5) +4[/tex]

       [tex]y' =m_2= 6[/tex]

Here m_1  and  m_2 are slops of the curve

Next we obtain the coordinates of the tangent lines

So  at [tex]x_1 = 3 [/tex]

   [tex]y_1 = 2(3)^2 + 4(3)[/tex]

  [tex]y_1 = 21[/tex]

So the coordinate for the first tangent line is  

    [tex](x_1 , y_1 ) = (3 , 21)[/tex]

At  [tex]x_2 = 0.5 [/tex]      

    [tex]y_2 = 2(0.5)^2 + 4(0.5)[/tex]

=>  [tex]y_2 = 2.5[/tex]

So the coordinate for the second  tangent line is  

    [tex](x_2 , y_2 ) = (0.5 , 2.5)[/tex]

Next we obtain the equation for the tangent lines

 So generally the slope is mathematically represented as

        [tex]m = \frac{y - y_1 }{x-x_1}[/tex]

For   [tex](x_1 , y_1 ) = (-3 , 21)[/tex] and  [tex]y' =m_1= 16[/tex]

       [tex] 16 = \frac{y -21 }{x-3)}[/tex]

=>    [tex]y = 16x - 27[/tex]

For  [tex](x_2 , y_2 ) = (0.5 , 2.5)[/tex] and  [tex]y' =m_2= 6[/tex]

       [tex] 6 = \frac{y -2.5 }{x-0.5}[/tex]

       [tex]y = 6x -0.5[/tex]

Generally the general expression for the slope of a line tangent to the curve of the function y=2x^2+4 at the point P(x,y) is mathematically evaluated by differentiating  y=2x^2+4 as follows

     [tex]y' = 4x[/tex]

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