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]
