calculator are allow for this part of the test review, please show me how to do this

Question

Answer 1

23) The function f(x) = - x⁴  + 20x²  - 20x - 100

This is a even degree polynomial function with negative leading coefficient.

Now graph the polynomial function.

1)Test points

Make the table of values to the polynomial.

Choose random values for x and find the corresponding values for y.

x	y = - x4  + 20x2  - 20x - 100	(x, y )
- 4	y = - (-4)4  + 20(-4)2  - 20(-4) - 100 = 44	(- 4, 44)
- 2	y = - (-2)4  + 20(-2)2  - 20(-2) - 100 = 4	(-2, 4)
0	y = - (0)4  + 20(0)2  - 20(0) - 100 = -100	(0, - 100)
1	y = - (1)4  + 20(1)2  - 20(1) - 100 = - 101	(1, - 101)
3	y = - (3)4  + 20(3)2  - 20(3) - 100 = - 61	(3, - 61)

2) End behavior y = - x⁴  + 20x²  - 20x - 100

Degree of polynomial is 4 and leading coefficient -1.

The graph of a polynomial function is always a smooth curve; that is, it has no breaks or corners.

All even degree polynomials behave on their ends like quadratics.

All even degree polynomials are either up on both ends and or down on both ends.depending on whether the polynomial has, respectively, a positive or negative leading coefficient.

The above polynomial even degree  polynomial with a negative leading coefficient .

So the graph down on both ends.

3)Graph

1.Draw a coordinate plane.

2.Plot the coordinate points found in the table.

3.Then sketch the graph, connecting the points with a smooth curve.

From the graph, we can observe the function maximum at (x, y) = (-3,388, 65.57)

The range of even degree polynomial with negative leading coefficient is (-∞, Ymax]

Range is (-∞, 65.57]

Range set is { y Є R: y ≤ 65.57}

Answer 2

24) The function

Apply derivative on each side with respect of x.

Apply quotient rule in derivatives d/dx(u/v) = [vu' - uv']/v²

u = x⁴ - 3, v = x³ + 3x² + 7

u' = 4x³ , v' = 3x² + 6x

.

Answer 3

25) The exponential function f(x) = 2^x

y = 2^x

Differentiating  on each side with respect of x .

Apply the formula d/dx(a^x) = a^x ln a

y' = 2^x ln 2

Substitute the values (x , y) = (5, 32) in above equation.

y' = 2⁵ ln 2

y' = 32 (0.6931)

y' = 22.1792

This is the slope of tangent line to the curve at (5, 32).

To find the tangent line equation, substitute the values of m = 22.1792 and (x, y ) = (5, 32) in the slope intercept form of an equation y = mx + b.

32 = 22.1792(5) + b

32 = 110.896 + b

b = - 78.896

Substitute m = 22.1792 and b = - 78.896 in y = mx + b.

Equation of tangent line is y = (22.1792)x - 78.896.