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High School Math : Quadratic Functions

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Example Questions

Example Question #1 : Quadratic Functions

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

Possible Answers:

Red line

None of them

Green line

Blue line

Purple 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 $\small {f(x)=ax^2+bx+c}$ .

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

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

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

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Example Question #2 : Quadratic Functions

Which of the following functions represents a parabola?

Possible Answers:

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

f(x) = 5

$f(x) = x^{2} - 9$

f(x) = x

$f(x) = x^{2} - y^{2}$

Correct answer:

$f(x) = x^{2} - 9$

Explanation:

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

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Example Question #1 : Center And Radius Of Circle Functions

Find the radius of the circle given by the equation:

x^2+4x+y^2-12y=24

Possible Answers:

Correct answer:

Explanation:

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

a^2x^2=x^2 , so a=1 .

-2abx=4x and a=1 , so -2b=4 and b=-2 .

Therefore, 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:

x^2+4x+y^2-12y=24

x^2+4x+ 4+y^2-12y=24+4

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

Now we do the same for :

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

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

We can now find :

r^2 = 64 = 8

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Example Question #2 : Center And Radius Of Circle Functions

Find the center of the circle given by the equation:

x^2+y^2-6x+18y=-65

Possible Answers:

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

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

$\left ( 5,5 \right )$

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

$\left ( 12,3 \right )$

Correct answer:

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

Explanation:

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

a^2x^2=x^2 , so a=1 .

-2abx=-6x and a=1 , so -2b=-6 and b=3 .

This gives 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:

x^2-6x+y^2+18y=-65

x^2-6x+9+y^2+18y=-65+9

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

Now we do the same for :

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

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

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

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Example Question #3 : Quadratic Functions

Find the -intercepts for the circle given by the equation:

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

Possible Answers:

$0\ and\ 1$

$0\ and\ 9$

$-2\ and\ 4$

$1\ and\ 9$

$-4\ and\ 2$

Correct answer:

$-2\ and\ 4$

Explanation:

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

(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:

$x-1=\sqrt{9}$ and $x-1=-\sqrt{9}$

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

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Example Question #4 : Quadratic Functions

Find the -intercepts for the circle given by the equation:

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

Possible Answers:

$-6\ and\ 12$

$-3\ and\ 5$

$3\ and\ -5$

$12\ and\ 6$

$8\ and\ 12$

Correct answer:

$-6\ and\ 12$

Explanation:

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

(x-3)^2+25=106

(x-3)^2=81

$x-3=\sqrt{81}=9$ and $x-3=-\sqrt{81}=-9$

We can then solve these two equations to obtain

$x=-6\ or\ 12$

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Example Question #5 : Quadratic Functions

Find the center and radius of the circle defined by the equation:

(x-3)^2+(y+5)^2=36

Possible Answers:

$\left ( -3,-5 \right )\ and\ r=18$

$\left ( -3,5 \right )\ and\ r=6$

$\left ( 3,5 \right )\ and\ r=6$

$\left ( -3,5 \right )\ and\ r=18$

$\left ( 3,-5 \right )\ and\ r=6$

Correct answer:

$\left ( 3,-5 \right )\ and\ r=6$

Explanation:

The equation of a circle is: $(x-x_{1})^2+(y-y_{1})^2=r^2$ where is the radius and $(x_{1},y_{1})$ is the center.

In this problem, the equation is already in the format required to determine center and radius. To find the -coordinate of the center, we must find the value of that makes (x-3)^2 equal to 0, which is 3. We do the same to find the y-coordinate of the center and find that y=5 . To find the radius we take the square root of the constant on the right side of the equation which is 6.

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Example Question #6 : Quadratic Functions

Find the center and radius of the circle defined by the equation:

$(x+12)^{2}+(y+10)^2=100$

Possible Answers:

$\left ( 10,-12 \right )\ and\ r=100$

$\left ( -12,10 \right )\ and\ r=25$

$\left ( 12,-10 \right )\ and\ r=10$

$\left ( -12,-10 \right )\ and\ r=10$

$\left ( 12,-10 \right )\ and\ r=100$

Correct answer:

$\left ( -12,-10 \right )\ and\ r=10$

Explanation:

The equation of a circle is: $(x-x_{1})^2+(y-y_{1})^2=r^2$ where is the radius and $(x_{1},y_{1})$ is the center.

In this problem, the equation is already in the format required to determine center and radius. To find the -coordinate of the center, we must find the value of that makes (x+12)^2 equal to , which is . We do the same to find the y-coordinate of the center and find that y=-10 . To find the radius we take the square root of the constant on the right side of the equation which is 10.

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High School Math : Quadratic Functions

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Quadratic Functions

Example Question #2 : Quadratic Functions

Example Question #1 : Center And Radius Of Circle Functions

Example Question #2 : Center And Radius Of Circle Functions

Example Question #3 : Quadratic Functions

Example Question #4 : Quadratic Functions

Example Question #5 : Quadratic Functions

Example Question #6 : Quadratic Functions

All High School Math Resources

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