Substitute the values of \(\displaystyle a=1, b=2,c=3, d=5\) into the expression.

\(\displaystyle \frac{a}{c}-\frac{b}{d} = \frac{1}{3}-\frac{2}{5}\)

In order to subtract these fractions, we will need a least common denominator.

Multiply the denominators together for the LCD.  Convert the two fractions.

\(\displaystyle \frac{1}{3}-\frac{2}{5} = \frac{1(5)}{3(5)}-\frac{2(3)}{5(3)} = \frac{5}{15}-\frac{6}{15}\)

Subtract the numerators now that the denominators are common.

The answer is:  \(\displaystyle -\frac{1}{15}\)

Given the expression \(\displaystyle a^3-b^2-c\) and the assigned values, substitute the values into the expression.

\(\displaystyle (-1)^3-(-3)^2-(-5)\)

Simplify this expression by order of operations.

\(\displaystyle -1-(9)+5 = -10+5 = -5\)

The answer is:   \(\displaystyle -5\)

Substitute the known values into the expression.

\(\displaystyle a^2-b^3 = (-6)^2-(-3)^3\)

Simplify the expression.

\(\displaystyle 36-(-27) = 36+27 = 63\)

The answer is:  \(\displaystyle 63\)

Substitute the assigned values into the expression.

\(\displaystyle (-1)^{(-2)}\times (-2)^{(-1)}\)

Simplify the negative exponents by rewriting both terms as fractions.

\(\displaystyle x^{-a} =\frac{1}{x^a}\)

\(\displaystyle (-1)^{(-2)}\times (-2)^{(-1)}= \frac{1}{(-1)^2} \times \frac{1}{(-2)^{1}}\)

Simplify the fractions.

\(\displaystyle 1\times \frac{1}{-2} = -\frac{1}{2}\)

The answer is:  \(\displaystyle -\frac{1}{2}\)

Substitute the known value of \(\displaystyle x=2, y=5\) into the equation.

\(\displaystyle 5= 3[2](A-7)\)

Simplify the equation.

\(\displaystyle 5=6(A-7)\)

Solve the right side by distribution.

\(\displaystyle 5=6A-42\)

Add 42 on both sides.

\(\displaystyle 5+42=6A-42+42\)

\(\displaystyle 47=6A\)

Divide by six on both sides.

\(\displaystyle \frac{47}{6}=\frac{6A}{6}\)

The answer is:  \(\displaystyle \frac{47}{6}\)

The term \(\displaystyle f(2) = 4\) means that \(\displaystyle f(x)=4\) when \(\displaystyle x=2\).

Substitute the terms in the function to solve for \(\displaystyle k\).

\(\displaystyle 4 = 5(2)-k\)

Solve for \(\displaystyle k\).

\(\displaystyle 4 = 10-k\)

Subtract 10 on both sides.

\(\displaystyle 4 -10= 10-k-10\)

\(\displaystyle -6 =-k\)

Divide by negative one to eliminate the negative signs.

\(\displaystyle \frac{-6 }{-1}=\frac{-k}{-1}\)

The answer is:  \(\displaystyle 6\)

The vertical asymptote(s) of the graph of a rational function such as \(\displaystyle f(x)\) can be found by evaluating the zeroes of the denominator after the rational expression is reduced. The expression is in simplest form, so set the denominator equal to 0 and solve for \(\displaystyle x\):

\(\displaystyle x^{2} - 16 = 0\)

Add 16 to both sides:

\(\displaystyle x^{2} - 16 + 16 = 0+ 16\)

\(\displaystyle x^{2} =16\)

Take the positive and negative square roots:

\(\displaystyle x = -\sqrt{16 } = -4\)

or 

\(\displaystyle x = \sqrt{16 } = 4\)

The graph of \(\displaystyle f(x)\) has two vertical asymptotes, the graphs of the lines \(\displaystyle x = -4\) and \(\displaystyle x= 4\).

\(\displaystyle g(f(x))\) is a composite function which means that the inside function is plugged into the outside function. So in this case, \(\displaystyle f(x)\) is plugged into \(\displaystyle g(x)\). In other words, replace the \(\displaystyle f(x)\) expression each time there is an \(\displaystyle x\) in the \(\displaystyle g(x)\) expression. 

In this case \(\displaystyle x-4\) would be plugged into each \(\displaystyle x\) in the \(\displaystyle g(x)\) expression. See below:

\(\displaystyle 5(x-4)+6(x-4)^2+3\)

This is then simplified to:

\(\displaystyle 5x-20+6x^2-48x+96+3\)

And then further simplified to:

\(\displaystyle 6x^2-43x+79\)

Almost all transformed functions can be written like this:

\(\displaystyle g(x)=a[f(b(x-c))]+d\)

where \(\displaystyle f(x)\) is the parent function. In this case, our parent function is \(\displaystyle f(x)=x^{2}\), so we can write \(\displaystyle g(x)\) this way:

\(\displaystyle g(x)=a[b(x-c)]^{2}+d\)

Luckily, for this problem, we only have to worry about \(\displaystyle a\) and \(\displaystyle d\).

\(\displaystyle a\) represents the vertical stretch factor of the graph.

• If \(\displaystyle |a|\) is less than 1, the graph has been vertically compressed by a factor of \(\displaystyle |a|\). It's almost as if someone squished the graph while their hands were on the top and bottom. This would make a parabola, for example, look wider.
• If \(\displaystyle |a|\) is greater than 1, the graph has been vertically stretched by a factor of \(\displaystyle |a|\). It's almost as if someone pulled on the graph while their hands were grasping the top and bottom. This would make a parabola, for example, look narrower.

\(\displaystyle d\) represents the vertical translation of the graph.

• If \(\displaystyle d\) is positive, the graph has been shifted up \(\displaystyle d\) units.
• If \(\displaystyle d\) is negative, the graph has been shifted down \(\displaystyle d\) units.

For this problem, \(\displaystyle a\) is 4 and \(\displaystyle d\) is -3, causing vertical stretch by a factor of 4 and a vertical translation down 3 units.

Begin with the standard equation for a parabola: \(\displaystyle f(x)=x^2\).

Horizontal shifts are represented by additions (leftward shifts) or subtractions (rightward shifts) within the parenthesis of the \(\displaystyle x\) term. To shift 4 units to the right, subtract 4 within the parenthesis.

\(\displaystyle f(x)=(x-4)^2\)

The width of the parabola is determined by the magnitude of the coefficient in front of \(\displaystyle x\). To make a parabola narrower, use a whole number coefficient. Halving the width indicates a coefficient of two.

\(\displaystyle f(x) = 2(x-4)^2\)