High School Math : Algebra II

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #812 : Algebra Ii

Which of the following is a vertical line? 

Possible Answers:

\(\displaystyle y = 0\)

\(\displaystyle y = 3x + 3\)

\(\displaystyle y = -x + 2\)

\(\displaystyle x = 2\)

\(\displaystyle y = 6\)

Correct answer:

\(\displaystyle x = 2\)

Explanation:

A vertical line has infinitely many values of \(\displaystyle y\) but only one value of \(\displaystyle x\). Thus, vertical lines are of the form \(\displaystyle x = c\), where \(\displaystyle c\) is a real number. The only equation of this form is \(\displaystyle x = 2\)

Example Question #1 : Linear Functions

Solve for the \(\displaystyle x\)- and \(\displaystyle y\)- intercepts:

\(\displaystyle 5x-2y=11\)

Possible Answers:

\(\displaystyle \left ( -\frac{11}{5},0\right)\ and \left ( 0,\frac{11}{2}\right)\)

\(\displaystyle \left ( \frac{11}{5},0\right)\ and \left ( 0,\frac{11}{2}\right)\)

\(\displaystyle \left ( -\frac{11}{5},0\right)\ and \left ( 0,-\frac{11}{2}\right)\)

\(\displaystyle \left ( \frac{11}{5},0\right)\ and \left ( 0,-\frac{11}{2}\right)\)

Correct answer:

\(\displaystyle \left ( \frac{11}{5},0\right)\ and \left ( 0,-\frac{11}{2}\right)\)

Explanation:

To solve for the \(\displaystyle x\)-intercept, set \(\displaystyle y\) to zero and solve for \(\displaystyle x\):

\(\displaystyle 5x-2y=11\)

\(\displaystyle 5x-2(0)=11\)

\(\displaystyle 5x=11\)

\(\displaystyle x=\frac{11}{5}\)

To solve for the \(\displaystyle y\)-intercept, set \(\displaystyle x\) to zero and solve for \(\displaystyle y\):

\(\displaystyle 5x-2y=11\)

\(\displaystyle 5(0)-2y=11\)

\(\displaystyle -2y=11\)

\(\displaystyle y=-\frac{11}{2}\)

Example Question #2 : Linear Functions

Solve:

\(\displaystyle x+3y=10\)

\(\displaystyle -2x+y=1\)

Possible Answers:

\(\displaystyle (1,3)\)

Infinitely many solutions

No solutions

\(\displaystyle (-2,1)\)

\(\displaystyle (3,-2)\)

Correct answer:

\(\displaystyle (1,3)\)

Explanation:

Use substution to solve this problem:

\(\displaystyle x+3y=10\) becomes \(\displaystyle x=10-3y\) and then is substituted into the second equation. Then solve for \(\displaystyle y\):

\(\displaystyle -2(10-3y)+y=1\), so \(\displaystyle y=3\) and \(\displaystyle x=1\) to give the solution \(\displaystyle (1,3)\).

Example Question #21 : Functions And Graphs

Write \(\displaystyle 2x+7y=8\) in slope-intercept form.

Possible Answers:

\(\displaystyle y=\frac{2}{7}x-\frac{8}{7}\)

\(\displaystyle y=\frac{-2}{7}x+\frac{8}{7}\)

\(\displaystyle y=\frac{-2}{7}x-\frac{8}{7}\)

\(\displaystyle y=\frac{2}{7}x+\frac{8}{7}\)

Correct answer:

\(\displaystyle y=\frac{-2}{7}x+\frac{8}{7}\)

Explanation:

Slope-intercept form is \(\displaystyle y=mx+b\).

\(\displaystyle 2x+7y=8\)

\(\displaystyle 7y=-2x+8\)

\(\displaystyle y=\frac{-2}{7}x+\frac{8}{7}\)

Example Question #1 : Graphing Parabolas

Based on the figure below, which line depicts a quadratic function?

Question_10

Possible Answers:

None of them

Red line

Blue line

Purple line

Green line

Correct answer:

Red line

Explanation:

A parabola is one example of a quadratic function, regardless of whether it points upwards or downwards.

The red line represents a quadratic function and will have a formula similar to \(\displaystyle \small {f(x)=ax^2+bx+c}\).

The blue line represents a linear function and will have a formula similar to \(\displaystyle \small {f(x)=ax+b\).

The green line represents an exponential function and will have a formula similar to \(\displaystyle \small {f(x)=ax^b}\).

The purple line represents an absolute value function and will have a formula similar to \(\displaystyle \small {f(x)=\left |ax+b \right |}.\).

Example Question #411 : Functions And Graphs

Which of the following functions represents a parabola?

Possible Answers:

\(\displaystyle f(x) = 5\)

\(\displaystyle f(x) = x\) 

\(\displaystyle f(x) = \frac{4x^{2}}{x^{3}}\)

\(\displaystyle f(x) = x^{2} - y^{2}\)

\(\displaystyle f(x) = x^{2} - 9\)

Correct answer:

\(\displaystyle f(x) = x^{2} - 9\)

Explanation:

A parabola is a curve that can be represented by a quadratic equation.  The only quadratic here is represented by the function \(\displaystyle f(x) = x^{2} - 9\), while the others represent straight lines, circles, and other curves.

Example Question #72 : Quadratic Functions

Find the radius of the circle given by the equation:

\(\displaystyle x^2+4x+y^2-12y=24\)

Possible Answers:

\(\displaystyle 9\)

\(\displaystyle 11\)

\(\displaystyle 10\)

\(\displaystyle 7\)

\(\displaystyle 8\)

Correct answer:

\(\displaystyle 8\)

Explanation:

To find the center or the radius of a circle, first put the equation in the standard form for a circle:  \(\displaystyle (x-x_{1})^2+(y-y_{1})^2=r^2\), where \(\displaystyle r\) is the radius and \(\displaystyle (x_{1},y_{1})\) is the center.

From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula \(\displaystyle (ax-b)^2=a^2x^2-2abx +b^2\) . 

\(\displaystyle a^2x^2=x^2\), so \(\displaystyle a=1\).

\(\displaystyle -2abx=4x\) and \(\displaystyle a=1\), so \(\displaystyle -2b=4\) and \(\displaystyle b=-2\).

Therefore, \(\displaystyle x^2+4x+4=(x+2)^2\).

Because the constant, in this case 4, was not in the original equation, we need to add it to both sides:

\(\displaystyle x^2+4x+y^2-12y=24\)

\(\displaystyle x^2+4x+ 4+y^2-12y=24+4\)

\(\displaystyle (x+2)^2+y^2-12y=28\)

Now we do the same for \(\displaystyle y\):

\(\displaystyle (x+2)^2+y^2-12y+36=28+36\)

\(\displaystyle (x+2)^2+(y-6)^2=64\)

We can now find \(\displaystyle r\):

\(\displaystyle r^2 = 64 = 8\)

Example Question #73 : Quadratic Functions

Find the center of the circle given by the equation:

\(\displaystyle x^2+y^2-6x+18y=-65\)

Possible Answers:

\(\displaystyle \left ( 3,-9 \right )\)

\(\displaystyle \left ( -8,5 \right )\)

\(\displaystyle \left ( 12,3 \right )\)

\(\displaystyle \left ( 5,5 \right )\)

\(\displaystyle \left ( 6,-18 \right )\)

Correct answer:

\(\displaystyle \left ( 3,-9 \right )\)

Explanation:

To find the center or the radius of a circle, first put the equation in standard form:  \(\displaystyle (x-x_{1})^2+(y-y_{1})^2=r^2\), where \(\displaystyle r\) is the radius and \(\displaystyle (x_{1},y_{1})\) is the center.

From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula \(\displaystyle (ax-b)^2=a^2x^2-2abx +b^2\) . 

\(\displaystyle a^2x^2=x^2\), so \(\displaystyle a=1\).

\(\displaystyle -2abx=-6x\) and \(\displaystyle a=1\), so \(\displaystyle -2b=-6\) and \(\displaystyle b=3\).

This gives \(\displaystyle x^2-6x+9=(x-3)^2\).

Because the constant, in this case 9, was not in the original equation, we must add it to both sides:

\(\displaystyle x^2-6x+y^2+18y=-65\)

\(\displaystyle x^2-6x+9+y^2+18y=-65+9\)

\(\displaystyle (x-3)^2+y^2+18y=-56\)

Now we do the same for \(\displaystyle y\):

\(\displaystyle (x-3)^2+y^2+18y+81=-56+81\)

\(\displaystyle (x-3)^2+(y+9)^2=25\)

We can now find the center: (3, -9)

Example Question #971 : Algebra Ii

Find the \(\displaystyle x\)-intercepts for the circle given by the equation:

\(\displaystyle (x-1)^2+y^2=9\)

Possible Answers:

\(\displaystyle -4\ and\ 2\)

\(\displaystyle 0\ and\ 9\)

\(\displaystyle 1\ and\ 9\)

\(\displaystyle 0\ and\ 1\)

\(\displaystyle -2\ and\ 4\)

Correct answer:

\(\displaystyle -2\ and\ 4\)

Explanation:

To find the \(\displaystyle x\)-intercepts (where the graph crosses the \(\displaystyle x\)-axis), we must set \(\displaystyle y=0\). This gives us the equation:

\(\displaystyle (x-1)^2=9\)

Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:

\(\displaystyle x-1=\sqrt{9}\) and \(\displaystyle x-1=-\sqrt{9}\)

We can then solve these two equations to obtain \(\displaystyle x=-2\ or\ 4\).

Example Question #1 : Circle Functions

Find the \(\displaystyle x\)-intercepts for the circle given by the equation:

\(\displaystyle (x-3)^2+(y+5)^2=106\)

Possible Answers:

\(\displaystyle 3\ and\ -5\)

\(\displaystyle 12\ and\ 6\)

\(\displaystyle -6\ and\ 12\)

\(\displaystyle -3\ and\ 5\)

\(\displaystyle 8\ and\ 12\)

Correct answer:

\(\displaystyle -6\ and\ 12\)

Explanation:

To find the \(\displaystyle x\)-intercepts (where the graph crosses the \(\displaystyle x\)-axis), we must set \(\displaystyle y=0\). This gives us the equation:

\(\displaystyle (x-3)^2+25=106\)

\(\displaystyle (x-3)^2=81\)

Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:

\(\displaystyle x-3=\sqrt{81}=9\) and \(\displaystyle x-3=-\sqrt{81}=-9\)

We can then solve these two equations to obtain

\(\displaystyle x=-6\ or\ 12\)

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