Question

find the point on the curve y=2x3-15x2+34x-20 where the tangent are parallel to y+2x​

Answer 1

Answer:

(2,-12)&(3,1)

Step-by-step explanation:

it is not easy to type the whole solution ,so I am explaining simply

derive the equation by x

then equate it to the slope (-2)

solve for x

we get x=3(or)2

substitute them in main equation

we get y

Answer 2

The points are (2,4) and (3,1).

Given:

• A curve $y = 2 {x}^{3} - 15 {x}^{2} + 34x - 20$
• A line $y + 2x = 0$

To find:

• Find the point on the curve where the tangent are parallel the given line.

Solution:

Formula/Concept to be used:

1. Slope of line y= mx+c is 'm'.
2. Slope of two parallel lines are equal.
3. Slope of curve is given by dy/dx.

Step 1:

Find the slope of curve.

Differentiate the curve with respect to x.

$\bf \frac{dy}{dx} = 6 {x}^{2} - 30 {x} + 34...eq1$

Step 2:

Find the slope of line.

$y = - 2x + 0$

Slope of line is -2.

Step 3:

Equate eq1 slope of line.

$\frac{dy}{dx} = - 2$

or

$6 {x}^{2} - 30 {x} + 34 = - 2$

or

$6 {x}^{2} - 30 {x} + 36 = 0$

or

$\bf {x}^{2} - 5x + 6 = 0...eq2$

Step 4:

Solve the quadratic equation in eq2.

Split the middle term.

${x}^{2} - 3x -2 x + 6 = 0$

or

$x(x - 3) - 2(x - 3) = 0$

or

$(x - 3)(x - 2) = 0$

or

$\bf x = 3$

or

$\bf x = 2$

Step 5:

Put the values of x and find the value of y.

Case 1: When x= 2

$y = 2 {(2)}^{3} - 15 {(2)}^{2} + 34(2)- 20$

or

$y = 16 - 60 + 68- 20$

or

$y = 84 - 80$

or

$\bf y = 4$

One point is (2,4).

Case 2: When x= 3

$y = 2 {(3)}^{3} - 15 {(3)}^{2} + 34(3)- 20$

or

$y = 54 - 135 +102 - 20$

or

$\bf y = 1$

The other point is (3,1)

Thus,

The points are (2,4) and (3,1) on the curve where tangents are parallel to the line y+2x=0.

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