How to graph a quadratic function

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Geometry › How to graph a quadratic function

Questions 1 - 10

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

Explanation

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

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

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

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

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

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

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

$f(x)= Ax^{2}+ Bx+ C$ has as its graph a vertical parabola on the coordinate plane. You are given that and , 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

I and III only

I and V only

I, II, and V only

I, III, and IV only

III and IV only

Explanation

I) The orientation of the parabola is determined solely by the sign of . Since , 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 ; by substitution, it can be found to be at . known to be equal to 9, so the -intercept can be determined to be .

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 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.

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

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

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

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

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

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

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

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

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

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

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

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

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

Explanation

The -intercepts, if any exist, can be found by setting :

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

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

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

The only -intercept is .

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 .

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

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

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

and

Explanation

The system of equations can be rewritten as

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

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

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

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

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

Give the -coordinate of the -intercept of the graph of the function

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

The graph of has no -intercept.

Explanation

The -intercept of the graph of is the point at which it intersects the -axis. Its -coordinate is 0; its -coordinate is , which can be found by substituting 0 for in the definition:

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

$f(0)= 0^{2}-3 \cdot 0-10 = 0 - 0 + 10 = 10$ ,

the correct choice.

Find the -intercept and range for the function:

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

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

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

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

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

Explanation

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

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

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

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

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

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

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

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

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

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

Find the equation based on the graph shown below:

Screen shot 2015 10 21 at 3.40.06 pm

Explanation

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

and the y-intercept is

These numbers are the solutions to the equation.

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

This would distribute to

and then simplify to

This also would show a y-intercept of .

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.

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

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

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

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

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

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

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

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

One point of intersection is .

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

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

The other point of intersection is .

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

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

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

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

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

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

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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