2x^2 -5x^2 -3x+7?

Question

(i) Find the equation of the tangent to the function at the point x = 2

ii) What is the value of y where this tangent crosses the y axis?
(iii) What is the value of x where this tangent crosses the x axis?

Answer 1

Your equation is incorrect. Correct it and perhaps I could help you with it. Your equation must have two variables whereby one of them is the subject of the equation. For example: y = 2x - 6.

Answer 2

2x^3 -5x^2 -3x +7 = y
The slope of the tangent to this curve at any point is given by the first derivative of y with respects to x. That is 6x^2 -10x -3. At x = 2, the gradient of the tangent is 24 - 20 - 3 = 1; The equation of the tangent, therefore, takes the form y = x + b . The y value for both the tangent and the above line must be equal when x = 2; Therefore, 16-20 -6 +7 = 2 + b; b = -5; The required tangent is y = x - 5
(ii) The value of x at the y axis is zero, so y = -5 at this point.
(iii) The value of y at the x axis is zero, so 0 = x - 5 or x = 5;

Answer 3

Let's assume you mean y = 2x^3-5x^2-3x+7

The tangent will have an equation of the form y=mx+c where m is the derivative of the original function at x=2.

The derivative of the function is dy/dx=6x^2-10x-3. When x=2 this is equal to 1, so m=1, ie y=x+c.

When x=2, the value of the function is y=-3, so by putting these values into y=x+c we get c=-5. So:

i) y=x-5

When the tangent crosses the y axis, x=0, so:

ii) y=-5 when the tangent crosses the y axis

When the tangent crosses the x axis, y=0, so:

iii) x=5 when the tangent crosses the x axis

Answer 4

I'm going to assume the first term is 2x^3.

The derivative of this is:
dy/dx = 6x^2 - 10x - 3, solving for x = 2 results in
6*2*2 - 10*2 - 3 = 24 - 20 - 3 = 1

So the slope of the tangent at x=2 is 1.

The original equation at x = 2 = 2*2*2*2 - 5*2*2 - 3*2 + 7 = 16 - 20 - 6 + 7 = -3

So we have a line whose slope is 1, and moves through the point (2, -3)

A line's formula is y = mx + b, and we have 3 of those terms:
-3 = 1*2 + b
-3 = 2 + b
b = -5, so the formula of the line (for part (i)) is y = 1x - 5

For part (ii), solve for x = 0 (the y-axis) y = 0 - 5, y = -5
For part (iii), solve for y = 0 (the x-axis) 0 = x - 5, x = 5

Ok?

Answer 5

(i) Assume f(x) = 2x³ - 5x² - 3x + 7

f `(x) = 6x² - 10x - 3

f ` (2) = 24 - 20 - 3 = 1 = gradient of tangent (m )
at x = 2

f(2) = 16 - 20 - 6 + 7 = -3

Tangent has m = 1 and passes thro` (2,-3)

Equation of tangent is given by:-
y -(-3) = 1 (x - 2)
y + 3 = x - 2
y = x - 5 is equation of tangent thro` (2,-3)

(ii) Cuts y axis when x = 0 ie when y = - 5

(iii) Cuts x axis when y = 0 ie when x = 5

Answer 6

(1)assume y=2x^3 -5x^2 -3x+7
let P(x1,y1)be a point on the
curve y=2x^3 -5x^2 -3x+7
at x=2,y=16-20-6+7= -3
the gradient of this curve is
(dy/dx)x1,where this symbol
denotes the value of dy/dx
when x1 is substituted for x
dy/dx=6x^2-10x-3
whenx1=2
(dx/dx)1= 1
hence,it's equation is;
y+3= (x-2)
>>y-x+5=0....(1)

(2)when the tangent crosses
the y-axis,x=0
substitute this value into (1)
y+5=0>>>>y= -5
therefore, y= -5 when the tangent
crosses the y-axis

(3)when the tangent crosses
the x-axis,y=0
substitute this value into (1)
0-x+5=0>>> x=5
therefore, x= 5 when the tangent
crosses the x-axis

i hope that this helps

Answer 7

Please check the question. Would it by any chance be 2x^3 at the start?

Answer 8

Please verify your question.