differentiation question...?

Question

a curve has the equation y=2x^3-5x+q, where q is a constant.
(i) if the tangent to the curve at the point where x=1 has y=intercept equal to 7, find the value of q.
(ii) find the equations of the normal to the curve which are parallel to the line x+y+5=0

pls show the steps

Answer 1

i) to find the tangent equation you first calculate the derivative of the function in x= 1... that is

y´(x) = 6x^2 -5 and y´(1) = 6(1^2) -5 = 1 this is de angular coefficient of the tangent line. For x=1 the point in the curve is

y(1) = 2 - 5 + q = q-3

The tangent equation is y- y1 = m(x-x1) where m is the angular coefficient and (x1,y1) is the point of the curve...
So, y- (q-3) = 1(x-1) or y = x -4 + q

The y-intercept is -4+q. As it is 7, we have -4+q = 7
and finally q = 11

ii) The tangent point is (1,8) and the tangent equation is
y = x +7
The normal at the point (1,8) is y -8 = -1(x-1) or
y = -x +9 ... this normal is parallel to the line x+y+5=0

I suppose, as you used the word "equations" the problem is in an arbitrary point (x1,y1)

If the normal must be parallel to x+y+5 = 0, then its angular coefficient must be m=-1
The equations of all normal to the curve, with the same angular coefficient are: y - y1 = -1(x - x1)
We can find only y1, substituting x1 in the curve equation:
y1 = 2x1^3 -5x1 + 11
And the equation becomes:
y -( 2x1^3 -5x1 + 11) = -1(x - x1)

and y = -x + 2x1^3 -4x1 + 11... are the equations of the normal... depending of x1

Answer 2

(i) We have y' = 6x^2 - 5. When x = 1, y' = 1 and y = 2 - 5 + q = -3 + q. Therefore the tangent line at (1,-3 + q) with slope 1 is y = x - 4 - q. The y-intercept is - 4 - q = 7, so q = -11.

(ii) The normals have the same slope as x + y + 5 = 0, which has slope -1. Thus, y' = 6x^2 - 5 = -1, so x = +/- sqrt(2/3). Find the corresponding points on the curve and write the equations of the lines through these points with slope -1.

Answer 3

Tangent to the line has the same slope of the lien at that point. Slope of line is the differential. y'=6x^2-5
In order to find q, we need to find y at x=1
We know the tangent line has the equation y=mx+b where:
y is what we're looking for.
m is the slope that we just found (6x^2-5)
b is the y-intercept givn as +7
Solve for y = {6(1)^2-5}*(1) + 7 = 8

Plug y=8 and x=1 into the original curve equation to find q:
ANSWER q=11