Calculus help!! 5 points please help me out :(?

Question

Q1) Find a parabola with equation y=ax^2+bx+c that has slope 3 at x=1 and slope −5 at x=−1 and passes through the point (3,17).

Q2) Find the value of c>0 such that the line y=x+16 is the tangent line to the curve c√x (i.e. intersects the curve at one point and shares the same slope at that point). c=?
Q3) Find the equation of the tangent line to:
a) y=x − 8x^2 at the point (2,−30) y=?
b) y=x^3 − 8x + 2 at the point (2,-6) y=?
Q4) If a ball is thrown vertically upwards with velocity 32 m/sec, its height (in metres) after t seconds is given by y=32t−4.9t^2

a) Find the velocity when t=2.

b) Find the value of t (to 4 decimal places) at which the ball is highest

Answer 1

(1).

The parabola equation is y = ax² + bx + c and slope 3 at x = 1.

To find the slope at x = 1, derivative with respect to x to each side.

dy/dx = 2ax + b.

3 = 2a + b  -----> equation 1.

 

The parabola equation is y = ax² + bx + c and slope - 5 at x = - 1.

To find the slope at x = - 1, derivative with respect to x to each side.

dy/dx = 2ax + b

- 5 = - 2a + b  -----> equation 2.

 

Solve the equation 1 and equation 2 through the elimination method

   3 =   2a + b  -----> equation 1.

- 5 = - 2a + b  -----> equation 2.

-----------------

- 2 = 2b ⇒ b = - 1.

To find the value of a, substitite the value of b = - 1 in either equation 1 or equation 2.

Equation 1 : 2a + b = 3 ⇒ 2a + (- 1) = 3 ⇒ 2a  = 4 ⇒ a  = 2.

 

The parabola y = ax² + bx + c passes through the point (3, 17).

17 = a(3)² + b(3) + c ⇒ 17 = 9a + 3b + c.

To find the value of c, substitute the value of a = 2 and b = - 1 in the equation 17 = 9a + 3b + c.

17 = 9(2) + 3(- 1) + c

17 = 18 - 3 + c

17 = 15 + c

2 = c.

 

The parabola equation is y = 2x² - x + 2.

Answer 2

(2).

The tangent line : y = x + 16 and curve : y = c√x.

The slope of y = x + 16 is 1.

To find the slope of curve y = c√x, derivative with respect to x to each side.

dy/dx = c/2√x

1 = c/2√x

c = 2√x.

 

Since intersects the curve at one point and shares the same slope at that point, c√x = x + 16.

(2√x)√x = x + 16

2x = x + 16

x = 16

Since c = 2√x ⇒ c = 2√(16) = 2(4) = 8.

 

The value of c = 8.

Answer 3

(3).(a).

The curve equation : y = x - 8x² and the point : (2, - 30)

dy/dx = 1 - 16x

m = 1 - 16(2) = 1 - 32 = - 31.

The point-slope form of line equation : y - y₁ = m(x - x₁), where m = slope and (x₁, y₁) = point.

Slope m = - 31 point (x₁, y₁) = (2, - 30).

y - (- 30) = (- 31)[ x - (2)]

y + 30 = (- 31)(x - 2)

y + 30 = - 31x + 62

y = - 31x + 32

The tangent line equation y = - 31x + 32.

 

(3).(b).

The curve equation : y = x³ - 8x + 2 and the point : (2, - 6)

dy/dx = 3x² - 8

m = 3(2)² - 8 = 12 - 8 = 4.

The point-slope form of line equation : y - y₁ = m(x - x₁), where m = slope and (x₁, y₁) = point.

Slope m = 4 point (x₁, y₁) = (2, - 6).

y - (- 6) = (4)[ x - (2)]

y + 6 = (4)(x - 2)

y + 6 = 4x - 8

y = 4x - 14

The tangent line equation y = 4x - 14.

 

Answer 4

(4).(a).

The equation : y = 32t - 4.9t² and the value of t = 2.

To find the velocity, derivative with respect to t.

dy/dt = 32 - 9.8t

v = 32 - 9.8(2) = 32 - 19.6 = 12.4 m/s.

The velocity is 12.4 m/s.

(4).(b).

At the highest ball, the velocity(dy/dt) is zero.

dy/dt = 32 - 9.8t

0 = 32 - 9.8t

9.8t = 32

t = 32 / 9.8 ≅ 3.2653 seconds

The value of t = 3.2653 seconds.