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Advanced Geometry : Graphing

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Example Questions

Example Question #166 : Coordinate Geometry

Which of the following is the equation of the line of symmetry of a horizontal parabola on the coordinate plane with its vertex at (8,4) ?

Possible Answers:

x= 8

x-2y = 0

x+y = 12

x-y = 4

y = 4

Correct answer:

y = 4

Explanation:

The line of symmetry of a horizontal parabola is a horizontal line, the equation of which takes the form y = b for some . The line of symmetry passes through the vertex, which here is (8,4) , so the equation must be y = 4 .

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Example Question #167 : Coordinate Geometry

The graphs of the functions f(x) and g(x) have the same -intercept.

If we define $f(x) = 2x^{2}+ 4x+7$ , which of the following is a possible definition of g(x) ?

Possible Answers:

$g(x) = -2x^{2}- 4x-14$

$g(x) = 6x^{2}+ 3x-7$

$g(x) =4x^{2}+ 16x+14$

$g(x) = x^{2}+ 2x+6$

$g(x) = 4x^{2}+ 8x+7$

Correct answer:

$g(x) = 4x^{2}+ 8x+7$

Explanation:

The -coordinate of the -intercept of the graph of a function of the form $f(x)= Ax^{2}+Bx+C$ - a quadratic function - is the point (0, f(0)) . Since $f(x)= A \cdot 0^{2}+B \cdot 0 +C = C$ , the -intercept is at the point (0,C) .

Because of this, the graph of $f(x) = 2x^{2}+ 4x+7$ has its -intercept at (0,7) . Among the other choices, only $g(x) = 4x^{2}+ 8x+7$ has a graph with its -intercept also at (0,7) .

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Example Question #71 : Coordinate Geometry

Find the vertex and determine if the vertex is a maximum or minimum for below.

$f(x)=-x^{2}+2x-8$

Possible Answers:

Vertex: (1,-5), Max

Vertex: (1,-7), Min

Vertex: (1,-7), Max

Vertex: (1,-5), Min

Correct answer:

Vertex: (1,-7), Max

Explanation:

The correct answer for the vertex is found by first finding the x of the vertex:

$x=\frac{-b}{2a}\ where\ a=-1\ and\ b=2$

Plug in a and b to get:

$x=\frac{-2}{2(-1)}=\boldsymbol{\mathit{}1}$

To find the y value of the vertex, plug in what was found for x above in the original f(x).

$f(x)=-x^{2}+2x-8\ plug\ in\ x=1$

$f(1)=-(1)^{2}+2(1)-8=\boldsymbol{\mathit{}-7}$

a common mistake here is the order of operations, at the beginning the 1 is squared before it is multipled by the negative out front.

$The\ vertex\ is\ \boldsymbol{\mathbf{\mathit{}}(1,-7)}$

Now we must consider if the vertex is a MAX or a MIN

Since the a value is negative, this means the parabola will open down, which means the vertex is the highest point on the graph.

$The\ vertex\ is\ the\ \boldsymbol{MAX}$

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Example Question #71 : Coordinate Geometry

What is the range for f(x) below?

$f(x)=-3(x-5)^{2}+7$

Possible Answers:

$y\leq5$

$y\geq 5$

$y\leq7$

$y\geq 7$

Correct answer:

$y\leq7$

Explanation:

$The\ -3\ at\ the\ beginning\ of\ the\ function\ determines\ if\ the\ graph\ opens\ up\ or\ down.$

$The\ parabola\ will\ open\ up\ if\ this\ value\ is\ positive.$

$The\ parabola\ will\ open\ down\ if\ this\ value\ is\ negative.$

$This\ parabola\ opens\ down\ which\ means\ the\ y-value\ of\ the\ vertex\ is\ the\ highest\ point.$

$This\ quadratic\ is\ in\ vertex\ form: f(x)=a(x-h)^{2}+k$

$The\ k\ value\ is\ the\ y-value\ of\ the\ vertex\ which\ is\ 7.$

$Given\ that\ the\ parabola\ opens\ down\ and\ the\ y-value\ of\ the\ vertex\ is\ 7,$

$the\ RANGE\ must\ be\ \mathbf{y\leq 7}.$

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Example Question #21 : How To Graph A Quadratic Function

Find the -intercept and range for the function:

$f(x)=5(x+3)^{2}-4.$

Possible Answers:

$Range: y\geq-4, y-int: (0,-4)$

$Range: y\geq-4, y-int: (0,41)$

$Range: y\leq -4, y-int: (0,41)$

$Range: y\leq -4, y-int: (0,-4)$

Correct answer:

$Range: y\geq-4, y-int: (0,41)$

Explanation:

$The\ range\ for\ a\ quadratic\ is\ determined\ from\ the\ y-value\ of\ the\ vertex.$

$This\ equation\ is\ written\ in\ vertex\ form: y=a(x-h)^{2}+k$

$The\ k -value\ is\ the\ y-coordinate\ for\ the\ vertex.$

$For\ this\ function\ y=5(x+3)^{2}-4,\ the\ k\ value\ is -4.$

$We\ can\ also\ see\ from\ the\ a-value,\ which\ is\ positive\ 5,\ that\ this\ parabola\ will\ open\ up\ on\ the\ graph.$

$\\Both\ of\ these\ together\ give\ us\ the\ range\ which\ is\ y\geq-4.\ It\ is \geq\ because\ the\ parabola\ opens\ up.$

$To\ find\ the\ y-intercept,\ you\ must\ plug\ in\ x=0\ to\ the\ original\ equation.$

$y=5(0+3)^{2}-4$

$Pay\ attention\ to\ order\ of\ operations\ by\ squaring\ the\ (-3)\ before\ multiplying\ by\ 5.$

$After\ simplifying\ we\ get\ y=41\ which\ is\ the\ y-intercept.$

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Example Question #72 : Coordinate Geometry

Find the equation based on the graph shown below:

Screen shot 2015 10 21 at 3.40.06 pm

Possible Answers:

y=x^2-7x+12

y=x^2-4

y=x^2+7x+12

y=x^2-x-12

y=x^2+x-12

Correct answer:

y=x^2-x-12

Explanation:

When you look at the graph, you will see the x-intercepts are

(4,0)(-3,0)

and the y-intercept is

(0,-12) .

These numbers are the solutions to the equation.

You can work backwards and see what the actual equation will come out as,

y=(x-4)(x+3) .

This would distribute to

y=x^2-4x+3x-12

and then simplify to

y=x^2-x-12 .

This also would show a y-intercept of (0,-12) .

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Example Question #164 : Advanced Geometry

$Find\ the\ vertex\ for\ the\ function: f(x)=4x^{2}-16x+11.$

Possible Answers:

(2,-5)

(0,11)

(-2,59)

(4,-16)

Correct answer:

(2,-5)

Explanation:

$Use \frac{-b}{2a}\ to\ calculate\ the\ x-value\ of\ the\ vertex.$

$\small Given\ the\ function\ f(x)=4x^{2}-16x+11\ is\ in\ standard\ form\ y=ax^{2}+bx+c$

$\small We\ can\ see\ that\ a=4, b=-16\ and\ c=11$

$\small Substitute\ these\ values\ in\ for\ x=\frac{-b}{2a}\ to\ get\ x=\frac{16}{8}=2$

$\small Now\ substitute\ x=2\ into\ the\ function\ f(x)\ to\ get:$

$\small f(2)=4(2)^{2}-16(2)+11=-5$

$\small The\ coordinates\ of\ the\ vertex\ are\ (2,-5)$

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Example Question #165 : Advanced Geometry

Determine the domain and range for the graph of the below function:

f(x)=-3x^2

Possible Answers:

$Domain:\mathbb{R}\and\ Range: y\geq-3$

$Domain:\mathbb{R}\and\ Range: y\leq0$

$Domain:\mathbb{R}\and\ Range: y\leq-3$

$Domain:\mathbb{R}\and\ Range: y\geq0$

Correct answer:

$Domain:\mathbb{R}\and\ Range: y\leq0$

Explanation:

When finding the domain and range of a quadratic function, we must first find the vertex.

$To\ find\ the\ vertex\ of\ a\ quadratic\ we\ must\ first\ find\ the\ x-coordinate\ using\ the\ formula:$

$x=\frac{-b}{2a}$

$We\ are\ given\ f(x)=-3x^2$

$This\ gives\ us\ an\ a-value\ of\ -3\ while\ b\ and\ c\ both\ equal\ 0.$

$Plugging\ these\ values\ in\ give\ us\ an\ x-coordinate\ and\ y-coordinate\ of\ 0.$

$The\ domain\ of\ all\ quadratics\ is\ ALL\ REAL\ NUMBERS\ (\mathbb{R})$

$With\ a\ negative\ a-value\ we\ know\ this\ quadratic\ opens\ down:$

$so\ the\ range\ is\ y\leq0$

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Example Question #166 : Advanced Geometry

Give the -coordinate(s) of the -intercept(s) of the graph of the function

$f(x)= x^{2}-3x-10$ .

Possible Answers:

-10, 1

-1, 10

-5, 2

The graph of f(x) has no -intercept.

-2, 5

Correct answer:

-2, 5

Explanation:

The -intercept(s) of the graph of f(x) are the point(s) at which it intersects the -axis. The -coordinate of each is 0; their -coordinate(s) are those value(s) of for which f(x) = 0 , so set up, and solve for , the equation:

f(x)= 0

$x^{2}-3x-10 = 0$

To solve this quadratic equation, first, factor the quadratic trinomial as

(x+a)(x+b)

by finding two integers with sum and product . From trial and error, we find the integers and 2, so the equation can be rewritten as

(x-5)(x+2) = 0

By the Zero Factor Principle, one of these two binomials is equal to 0, so either

x-5 = 0 ,

in which case

x = 5 ,

x+2 = 0 ,

in which case

x= -2 .

The graph has two -intercepts at the points (-2, 0) and (5, 0) .

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Example Question #71 : Graphing

Give the domain of the function $f(x) = x^{2} + 4x - 17$ .

Possible Answers:

$\left \{x| x \ge -2\right \}$

$\left \{ x | x > -17 \right \}$

$\left \{ x | x >0 \right \}$

$\left \{ x | x > -21 \right \}$

The set of all real numbers

Correct answer:

The set of all real numbers

Explanation:

f(x) is a polynomial function, and as such has the set of all real numbers as its domain.

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Advanced Geometry : Graphing

Study concepts, example questions & explanations for Advanced Geometry

All Advanced Geometry Resources

Example Questions

Example Question #166 : Coordinate Geometry

Example Question #167 : Coordinate Geometry

Example Question #71 : Coordinate Geometry

Example Question #71 : Coordinate Geometry

Example Question #21 : How To Graph A Quadratic Function

Example Question #72 : Coordinate Geometry

Example Question #164 : Advanced Geometry

Example Question #165 : Advanced Geometry

Example Question #166 : Advanced Geometry

Example Question #71 : Graphing

All Advanced Geometry Resources

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