final pratice exam help please no calculator

Question

Answer 1

21)

General form of conic section Ax² + Cy² + Dx + Ey + F = 0

Conics can be classified according to the coefficients (x² and y²) of the equation.

If the coefficients of  x² and y² Are of the same sign but unequal coefficients, then the graph will be an ellipse.

If both coefficients of x² and y² Are of the same sign with unequal coefficients then the graph will be circle.

If coefficients are of opposite sign, then it will be a hyperbola.

A parabola results when there is an " x² " but no " y² " (or) a " y² "  but no "x² ".

In the options, there is one equation having " x² " but no " y² ".

Therefore, x² - 2x + 3y = 1 graph represents a parabola in the xy - plane.

Option (b) is correct.

Answer 2

22) The function f(x) = x² + 3

To f(x + h), substitute (x + h) for x in f(x) = x² + 3 .

f(x +  h) = (x + h)² + 3

f(x +  h) = x² + h² + 2xh + 3

[f(x + h) - f(x)]/h = [(x² + h² + 2xh + 3) - (x² + 3)]/h

= (x² + h² + 2xh + 3 - x² - 3)/h

= (h² + 2xh)/h

= h(h + 2x)/h

= h + 2x

[f(x + h) - f(x)]/h = 2x + h

Option (c) is correct.

Answer 3

23) The function  f(x)= 3x⁴  - 4x³  + 5

y = 3x⁴  - 4x³  + 5

Differentiating with respect to x.

y' = 12x³  - 12x²

Here y' is slope of the tangent line.

Slope of the horizontal line is zero.So equate first derivative = 0

12x³  - 12x²  = 0

12x²(x - 1) = 0

Apply zero product property.

12x² = 0 and x - 1 = 0

x = 0 and x = 1

Using the x - value find the corresponding y - value with the curve.

y = 3(0)⁴ - 4(0)³  + 5 = 5

The point ( 0, 5)

For x = 1,

y = 3(1)⁴  - 4(1)³  + 5

= 3 - 4 + 5 = 4

The horizontal tangent points are (0, 5) and (1, 4).

Option (a) is correct.