Find an equation of the line tangent to the graph of y= x^2 + sin(pi/2)x at x=-1

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Let the equation for the line be :

y-y1 = m(x-x1)

where (x1,y1) is any point on the line and m is the slope.

We know that the slope is the derivative at -1

Then we will calculate y'

y= x^2 + sin(pi/2)x

==> y' = 2x + (pi/2)cos(pi/2)x

No substitute with x= -1

==>y' = m = 2*-1 + (pi/2) cos -piu/2

= -2+ 0 = -2

Then the slope m= -2

Now sine the line passes through x= -1, then we will find y value:

y= x^2 + sin(pi/2)x

= -1^2 - sin(pi/2)

= 1 - 1= 0

Then th line passes though the poin (-1,0)

Now we will substitute:

y- 0 = -2(x--1)

==>y= -2(x+1)

**==> y= -2x - 2**

To find the tangent to the curve y = x^2+sin(pi/2)x at x = 1.

At x= 1, y = = 1^2 +sin (pi/2)(1) = 1 +1 = 2.

So (x,y ) = (1,2) is a point on the curve where we like to determine the equation of the tangent to the curve.

We know the tangent at (x1 y1) is given by:

y-y1 ={ (dy/dx) at (x1 , y1) }(x-x1))....(1)

We differentiate y = x^2+sin (pi/2)x.

dy/dx = 2x+ (pi/2)cos(pi/2)x

{dy/dx at x = 1} = 2+(pi/2) cos(pi/2) = 2+(pi/2)*0 = 2.

Therefore substituting (x1 ,y1) = (1,2) and {dy/dx ) at x= 1) = 2 in (1), we get the equation of the tangent:

y - 2 = 2(x-1).

y-2 = 2x-2 .

y = 2x is the equation of the tangent to y= x^2+sin (pi/2)x at x= 1.

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