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 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.

Answer 1

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 $x = 3$ 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) ​?

Answer:

The  generally expression for the slope of $y = 2x^2 + 4x$ is  $y' = 4x +4$

The graph is shown on the first uploaded image

The  generally expression for the slope of $y=2x^2+4$ is   $y' = 4x$

Step-by-step explanation:

From the question we are told that

  The  equation of the curve is $y = 2x^2 + 4x$

First we differentiate the equation

So  

     $y' = 4x +4$

Therefore the generally expression for the slope tangent to the curve $y=2x^2+4x$ is   $y' = 4x +4$

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

So  for $x_1 = 3$

    $y' = 4(3) +4$

     $y' =m_1= 16$

And  for  $x_2 = 0.5$

      $y' = 4(0.5) +4$

       $y' =m_2= 6$

Here m_1  and  m_2 are slops of the curve

Next we obtain the coordinates of the tangent lines

So  at $x_1 = 3$

   $y_1 = 2(3)^2 + 4(3)$

  $y_1 = 21$

So the coordinate for the first tangent line is  

    $(x_1 , y_1 ) = (3 , 21)$

At  $x_2 = 0.5$      

    $y_2 = 2(0.5)^2 + 4(0.5)$

=>  $y_2 = 2.5$

So the coordinate for the second  tangent line is  

    $(x_2 , y_2 ) = (0.5 , 2.5)$

Next we obtain the equation for the tangent lines

 So generally the slope is mathematically represented as

        $m = \frac{y - y_1 }{x-x_1}$

For   $(x_1 , y_1 ) = (-3 , 21)$ and  $y' =m_1= 16$

       $16 = \frac{y -21 }{x-3)}$

=>    $y = 16x - 27$

For  $(x_2 , y_2 ) = (0.5 , 2.5)$ and  $y' =m_2= 6$

       $6 = \frac{y -2.5 }{x-0.5}$

       $y = 6x -0.5$

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

     $y' = 4x$