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Advanced Geometry : How to graph a quadratic function

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Example Questions

Example Question #1 : Graphing A Quadratic Function

What are the possible values of if the parabola of the quadratic function $f (x) = Ax^{2} + 4x- 2$ is concave upward and does not intersect the -axis?

Possible Answers:

A > 4

-2 < A < 0

-2 < A < 4

The parabola cannot exist for any value of .

A > -2

Correct answer:

The parabola cannot exist for any value of .

Explanation:

If the graph of $f (x) = Ax^{2} + 4x- 2$ is concave upward, then A > 0 .

If the graph does not intersect the -axis, then $Ax^{2} + 4x- 2 = 0$ has no real solution, and the discriminant $4^{2} - 4 \cdot A \cdot \left ( -2\right )$ is negative:

$4^{2} - 4 \cdot A \cdot \left ( -2\right ) < 0$

16 +8 A < 0

16 +8 A - 16 < 0 - 16

8 A < - 16

$8 A \div 8< - 16 \div 8$

A < -2

For the parabola to have both characteristics, it must be true that A < -2 and A > 0 , but these two events are mutually exclusive. Therefore, the parabola cannot exist.

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Example Question #2 : Graphing A Quadratic Function

Which of the following equations has as its graph a vertical parabola with line of symmetry x = 7 ?

Possible Answers:

$f(x)=7x^{2}-49x+140$

$f(x)= 7x^{2}+49x+140$

$f(x)=7x^{2}-98x+140$

$f(x)=7x^{2} +140$

$f(x)=7x^{2}+98x+140$

Correct answer:

$f(x)=7x^{2}-98x+140$

Explanation:

The graph of $f(x)= Ax^{2}+Bx+ C$ has as its line of symmetry the vertical line of the equation

$x=- \frac{B}{2A}$

Since A=7 in each choice, we want to find such that

$7=- \frac{B}{2A} = - \frac{B}{2(7)} = - \frac{B}{14}$

$- \frac{B}{14} = 7$

$B = 7\cdot (-14)=-98$

so the correct choice is $f(x)=7x^{2}-98x+140$ .

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Example Question #141 : Advanced Geometry

Which of the following equations has as its graph a concave-right horizontal parabola?

Possible Answers:

$y=- 7x^{2}- 4x+13$

$x=7y^{2}- 4y+13$

$x=- 7y^{2}- 4y+13$

None of the other responses gives a correct answer.

$y= 7x^{2}- 4x+13$

Correct answer:

$x=7y^{2}- 4y+13$

Explanation:

A horizontal parabola has as its equation, in standard form,

$x= Ay^{2}+ By+ C$ ,

with A,B,C real, nonzero.

Its orientation depends on the sign of . In the equation of a concave-right parabola, is positive, so the correct choice is $x=7y^{2}- 4y+13$ .

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Example Question #2 : Graphing A Quadratic Function

The graphs of the functions f(x) and g(x) have the same line of symmetry.

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

Possible Answers:

None of the other responses gives a correct answer.

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

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

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

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

Correct answer:

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

Explanation:

The graph of a function of the form $f(x)= Ax^{2}+Bx+C$ - a quadratic function - is a vertical parabola with line of symmetry $x= - \frac{B}{2A}$ .

The graph of the function $f(x) = 2x^{2}+ 4x+7$ therefore has line of symmetry

$x= - \frac{4}{2 (2)}$ , or x = -1

We examine all four definitions of to find one with this line of symmetry.

$g(x) = 2x^{2}-4x-7$ :

$x= - \frac{-4}{2 (2)}$ , or x = 1

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

$x= - \frac{ 4}{2 (1)}$ , or x = -2

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

$x= - \frac{ 8}{2 (2)}$ , or x = -2

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

$x= - \frac{ 8}{2 (4)}$ , or x = -1

Since the graph of the function $g(x) = 4x^{2}+8x+7$ has the same line of symmetry as that of the function $f(x) = 2x^{2}+ 4x+7$ , that is the correct choice.

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Example Question #5 : Graphing A Quadratic Function

Give the -coordinate of a point at which the graphs of the equations

$y= x^{2}+ 2x- 7$

and

$y = 2x^{2}- 6x+5$

intersect.

Possible Answers:

Correct answer:

Explanation:

We can set the two quadratic expressions equal to each other and solve for .

$y= x^{2}+ 2x- 7$ and $y = 2x^{2}- 6x+5$ , so

$2x^{2}- 6x+5 = x^{2}+ 2x- 7$

$2x^{2}- 6x+5 -( x^{2}+ 2x- 7)= x^{2}+ 2x- 7 -( x^{2}+ 2x- 7)$

$2x^{2}- 6x+5 - x^{2}- 2x+ 7=0$

$x^{2}- 8x+12=0$

(x-2)(x-6) = 0

The -coordinates of the points of intersection are 2 and 6. To find the -coordinates, substitute in either equation:

$y= x^{2}+ 2x- 7$

$y= 2^{2}+ 2 \cdot 2 - 7$

y= 4+4 - 7

y=1

One point of intersection is (2,1) .

$y= x^{2}+ 2x- 7$

$y= 6^{2}+ 2 \cdot 6 - 7$

y= 36+ 12 - 7

y = 41

The other point of intersection is (6,41) .

1 is not among the choices, but 41 is, so this is the correct response.

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Example Question #51 : Coordinate Geometry

Give the set of intercepts of the graph of the function $f(x)= \frac{1}{7} (x-7)^{2}$ .

Possible Answers:

$\left \{ (-7,0), (0,7) \right \}$

$\left \{ (7,0), (0,-7),(0,7) \right \}$

$\left \{ (7,0), (0,7) \right \}$

$\left \{ (7,0), (0,-7) \right \}$

$\left \{ (-7,0), (0,-7),(0,7) \right \}$

Correct answer:

$\left \{ (7,0), (0,7) \right \}$

Explanation:

The -intercepts, if any exist, can be found by setting f(x) = 0 :

$f(x)= \frac{1}{7} (x-7)^{2} = 0$

$\frac{1}{7} (x-7)^{2} \cdot 7 = 0\cdot 7$

$(x-7)^{2} = 0$

x-7 = 0

x= 7

The only -intercept is (7,0) .

The -intercept can be found by substituting 0 for :

$f(0)= \frac{1}{7} (0-7)^{2} = \frac{1}{7} ( -7)^{2}= \frac{1}{7} \cdot 49 = 7$

The -intercept is (0,7) .

The correct set of intercepts is $\left \{ (7,0), (0,7) \right \}$ .

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Example Question #7 : Graphing A Quadratic Function

Give the -coordinate of a point of intersection of the graphs of the functions

$f(x) = x^{2}- 4x+ 7$

and

g(x) = 4x .

Possible Answers:

Correct answer:

Explanation:

The system of equations can be rewritten as

$y= x^{2}- 4x+ 7$

y = 4x .

We can set the two expressions in equal to each other and solve:

$x^{2}- 4x+ 7 = 4x$

$x^{2}- 8x+ 7 =0$

(x-1)(x-7)= 0

x= 1,x=7

We can substitute back into the equation y = 4x , and see that either y = 4 or y = 28 . The latter value is the correct choice.

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Example Question #8 : Graphing A Quadratic Function

$f(x)= Ax^{2}+ Bx+ C$ has as its graph a vertical parabola on the coordinate plane. You are given that A= - 5 and C = 9 , but you are not given .

Which of the following can you determine without knowing the value of ?

I) Whether the graph is concave upward or concave downward

II) The location of the vertex

III) The location of the -intercept

IV) The locations of the -intercepts, if there are any

V) The equation of the line of symmetry

Possible Answers:

I and V only

I and III only

III and IV only

I, II, and V only

I, III, and IV only

Correct answer:

I and III only

Explanation:

I) The orientation of the parabola is determined solely by the sign of . Since A= -5 < 0 , the parabola can be determined to be concave downward.

II and V) The -coordinate of the vertex is $-\frac{B}{2A}$ ; since you are not given , you cannot find this. Also, since the line of symmetry has equation $x=-\frac{B}{2A}$ , for the same reason, you cannot find this either.

III) The -intercept is the point at which x = 0 ; by substitution, it can be found to be at (0,C) . known to be equal to 9, so the -intercept can be determined to be (0,9) .

IV) The -intercept(s), if any, are the point(s) at which $f(x)= Ax^{2}+ Bx+ C = 0$ . This is solvable using the quadratic formula

$x= \frac{-B \pm \sqrt{B^{2}-4AC}}{2A}$

Since all three of A,B, and must be known for this to be evaluated, and only is known, the -intercept(s) cannot be identified.

The correct response is I and III only.

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Example Question #161 : Coordinate Geometry

Which of the following equations can be graphed with a vertical parabola with exactly one -intercept?

Possible Answers:

$f(x)= 3x^{2}- 12x + 15$

$f(x)= 3x^{2}- 12x + 12$

$f(x)= 3x^{2}- 12x + 6$

$f(x)= 3x^{2}- 12x + 9$

$f(x)= 3x^{2}- 12x + 3$

Correct answer:

$f(x)= 3x^{2}- 12x + 12$

Explanation:

The graph of $f(x)= Ax^{2}+Bx+ C$ has exactly one -intercept if and only if

$Ax^{2}+Bx+ C = 0$

has exactly one solution - or equivalently, if and only if

$B^{2}-4AC = 0$

Since in all three equations, A = 3,B = -12 , we find the value of that makes this statement true by substituting and solving:

$(12)^{2}-4(3)C = 0$

144-12C = 0

12C = 144

C= 12

The correct choice is $f(x)= 3x^{2}- 12x + 12$ .

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Example Question #61 : Graphing

$f(x)= Ax^{2}+ Bx+ C$ has as its graph a vertical parabola on the coordinate plane. You are given that A= 6 , but you are given neither nor .

Which of the following can you determine without knowing the values of and ?

I) Whether the curve opens upward or opens downward

II) The location of the vertex

III) The location of the -intercept

IV) The locations of the -intercepts, if there are any

V) The equation of the line of symmetry

Possible Answers:

V only

III only

III and IV only

I only

II and V only

Correct answer:

I only

Explanation:

I) The orientation of the parabola is determined solely by the value of . Since A= 6 > 0 , the parabola can be determined to open upward.

III) The -intercept is the point at which x = 0 ; by substitution, it can be found to be at (0,C) . is unknown, so the -intercept cannot be found.

IV) The -intercept(s), if any, are the point(s) at which $f(x)= Ax^{2}+ Bx+ C = 0$ . This is solvable using the quadratic formula

$x= \frac{-B \pm \sqrt{B^{2}-4AC}}{2A}$

Since all three of A,B, and must be known for this to be evaluated, and only is known, the -intercept(s) cannot be identified.

The correct response is I only.

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Advanced Geometry : How to graph a quadratic function

Study concepts, example questions & explanations for Advanced Geometry

All Advanced Geometry Resources

Example Questions

Example Question #1 : Graphing A Quadratic Function

Example Question #2 : Graphing A Quadratic Function

Example Question #141 : Advanced Geometry

Example Question #2 : Graphing A Quadratic Function

Example Question #5 : Graphing A Quadratic Function

Example Question #51 : Coordinate Geometry

Example Question #7 : Graphing A Quadratic Function

Example Question #8 : Graphing A Quadratic Function

Example Question #161 : Coordinate Geometry

Example Question #61 : Graphing

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