Question

Please use the graph above to answer these questions
1) Show that a = -4, b = -11, and c = 30

2) Determine the co-ordinates of A and B

3) Determine the equation of line g

4) Determine the co-ordinates of C

5) Determine the gradient of the tangent to f at its point of inflection

Answer 1

Answer:

1) f(x) = (x + 3)(x - 2)(x - 5)

= (x + 3)(x² - 7x + 10)

= x³ - 7x² + 10x + 3x² - 21x + 30

= x³ - 4x² - 11x + 30

So a = -4, b = -11, and c = 30.

2) f'(x) = 3x² - 8x - 11

3x² - 8x - 11 = 0

(x + 1)(3x - 11) = 0

x = -1, 3 2/3

f(-1) = 2(-3)(-6) = 36

A is at (-1, 36).

f(11/3) = -400/27

B is at (11/3, -400/27)

= (3 2/3, -14 22/27).

3) f(0) = 30, f'(0) = -11

g(x) = -11x + 30

4) x³ - 4x² - 11x + 30 = -11x + 30

x³ - 4x² = 0

x²(x - 4) = 0

x = 4--->f(4) = 7(2)(-1) = -14

C is at (4, -14).

5) f''(x) = 6x - 8

6x - 8 = 0

x = 4/3 = 1 1/3

f'(4/3) = 3(4/3)² - 8(4/3) - 11

= 16/3 - 32/3 - 11

= -49/3

Answer 2

Answer:

$\textsf{1)}\quad f(x)=x^3-4x^2-11x+30$

$\textsf{2)}\quad A (-1, 36),\;\;B\left(\dfrac{11}{3}, -\dfrac{400}{27}\right)$

$\textsf{3)}\quad g(x)=-11x+30$

$\textsf{4)}\quad C(4, -14)$

$\textsf{5)}\quad -\dfrac{49}{3}$

Step-by-step explanation:

Question 1

To find the coefficients a, b and c of function f(x) = x³ + ax² + bx + c, we can use its x-intercepts (-3, 0), (2, 0) and (5, 0) to write the function in its factored form.

The factored form of a cubic function can be expressed as:

$f(x) = a(x - r_1)(x - r_2)(x - r_3)$

where 'a' is the leading coefficient and rₙ are the roots (or zeros) of the cubic function.

Given that the leading coefficient of f(x) is 1, and its roots are x = -3, x = 2 and x = 5, then:

$f(x) = 1(x - (-3))(x - 2)(x - 5)\\\\f(x) = (x+3)(x - 2)(x - 5)$

Expand:

$f(x)=(x^2+x-6)(x-5)\\\\f(x)=x^3-5x^2+x^2-5x-6x+30\\\\f(x)=x^3-4x^2-11x+30$

Comparing f(x) = x³ - 4x² - 11x + 30 this f(x) = x³ + ax² + bx + c, the coefficients a, b and c are:

$\Large\boxed{\boxed{a=-4,\;\;b=-11,\;\;c=30}}$

$\dotfill$

Question 2

The gradient of the tangent line to the graph at its turning points is zero, meaning the derivative of the function at this point is zero.

To find the x-coordinates of the turning points A and B, set the derivative of function f(x) to zero and solve for x.

Differentiate f(x):

$f'(x)=3x^2-8x-11$

Set f'(x) to zero and solve for x:

$3x^2-8x-11=0\\\\3x^2-8x-11=0\\\\3x^2+3x-11x-11=0\\\\3x(x+1)-11(x+1)=0\\\\(3x-11)(x+1)=0\\\\\\(x+1)=0 \implies x=-1\\\\(3x-11)=0 \implies x=\dfrac{11}{3}$

Therefore, the x-coordinates of points A and B are x = -1 and x = 11/3, respectively.

To find the corresponding y-coordinates, substitute the x-coordinates into f(x).

Substitute x = -1 into f(x):

$y_A=f(-1)\\\\y_A=(-1)^3-4(-1)^2-11(-1)+30\\\\y_A=-1-4+11+30\\\\y_A=36$

Therefore, the coordinates of point A are:

$\Large\boxed{\boxed{A (-1, 36)}}$

Substitute x = 11/3 into f(x):

$y_B=f\left(\dfrac{11}{3}\right)\\\\\\y_B=\left(\dfrac{11}{3}\right)^3-4\left(\dfrac{11}{3}\right)^2-11\left(\dfrac{11}{3}\right)+30\\\\\\y_B=-\dfrac{400}{27}$

Therefore, the coordinates of point B are:

$\Large\boxed{\boxed{B\left(\dfrac{11}{3}, -\dfrac{400}{27}\right)}}$

$\dotfill$

Question 3

Function g(x) = mx + c is tangent to f(x) at point D.

Given that point D is the y-intercept of functions f(x) and g(x), and function f(x) intersects the y-axis at y = 30, then the y-intercept (c) of function g(x) is also 30:

$g(x) = mx + 30$

Given that g(x) is tangent to f(x) at point D, we can find the gradient (m) of function g(x) by substituting the x-coordinate of point D into f'(x):

$m=f'(0)\\\\m=3(0)^2-8(0)-11\\\\m=-11$

So, the gradient of g(x) is m = -11.

Therefore, the equation of line g is:

$\Large\boxed{\boxed{g(x)=-11x+30}}$

$\dotfill$

Question 4

Point C is the second point of intersection of f(x) and g(x). To find the x-coordinate of point C, set the two functions equal to each other and solve for x:

$f(x)=g(x)\\\\x^3-4x^2-11x+30=-11x+30\\\\x^3-4x^2=0\\\\x^2(x-4)=0\\\\\\x^2=0 \implies x=0\\\\(x-4)=0 \implies x=4$

As the x-coordinate of point D is x = 0, the x-coordinate of point C must be x = 4. To find the corresponding y-coordinate, substitute x = 4 into g(x):

$y_C=g(4)\\\\y_C=-11(4)+30\\\\y_C=-14$

Therefore, the coordinates of point C are:

$\Large\boxed{\boxed{C(4, -14)}}$

$\dotfill$

Question 5

At a point of inflection, the second derivative equals zero.

Differentiate f'(x) to find f''(x):

$f''(x)=6x-8$

Set f''(x) to zero and solve for x:

$f''(x)=0\\\\\\6x-8=0\\\\\\6x=8\\\\\\x=\dfrac{8}{6}\\\\\\x=\dfrac{4}{3}$

To find the gradient of the tangent to f(x) at its point of inflection, substitute x = 4/3 into f'(x):

$f'\left(\dfrac{4}{3}\right)=3\left(\dfrac{4}{3}\right)^2-8\left(\dfrac{4}{3}\right)-11\\\\\\f'\left(\dfrac{4}{3}\right)=-\dfrac{49}{3}$

Therefore, the gradient of the tangent to f(x) at its point of inflection is:

$\Large\boxed{\boxed{-\dfrac{49}{3}}}$