Is it possible to find the vertex of the parabola using the equation -b/2a as well as the other equations listed in the article?

yes, it is possible, you will need to use -b/2a for the x coordinate of the vertex and another formula k=c- b^2/4a for the y coordinate of the vertex. good luck, hope this helped

How do you get the formula from looking at the parabola? I am having trouble when I try to work backward with what he said.

You can get the formula from looking at the graph of a parabola in two ways: Either by considering the roots of the parabola or the vertex. A parabola is not like a straight line that you can find the equation of if you have two points on the graph, because there are multiple different parabolas that can go through a given set of two points. Instead you need three points, or the vertex and a point. In the last practice problem on this article, you're asked to find the equation of a parabola. Think about how you can find the roots of a quadratic equation by factoring. The same principle applies here, just in reverse. You can figure out the roots (x-intercepts) from the graph, and just put them together as factors to make an equation. In this form, the equation for a parabola would look like y = a(x - m)(x - n). "a" is a coefficient (responsible for vertically stretching/flipping the parabola and thus doesn't affect the roots), and the roots of the graph are at x = m and x = n. Because the graph in the problem has roots at 3 and -1, our equation would look like y = a(x + 1)(x - 3). The only one that fits this is answer choice B), which has "a" be -1. You can also find the equation of a quadratic equation by finding the coordinates of the vertex from a graph, then plugging that into vertex form, and then picking a point on the parabola to use in order to solve for your "a" value. Thirdly, I guess you could also use three separate points to put in a system of three equations, which would let you solve for the "a", "b", and "c" in the standard form of a quadratic, but that's too much work for the SAT.

my sat is on 13 of march(probably after5 days ) n i'm craming over maths I just need 500 to 600 score for math so which topics should I focus on more ??

Also, remember not to stress out over it. Make sure to get a full nights. Good luck on your exam!

how would i graph this though f(x)=2(x-3)^2-2

Following the steps in the article, you would graph this function by following the steps to transform the parent function of y = x^2. Here, we see that 3 is subtracted from x inside the parentheses, which means that we translate right by 3. We subtract 2 from the final answer, so we move down by 2. Our vertex will then be right 3 and down 2 from the normal vertex (0,0), at (3, -2). From here, we see that there's a coefficient outside the parentheses, which means we vertically stretch the function by a factor of 2. The easiest way to graph this would be to find the vertex and direction that it opens, and then plug in a point for x and see what you get for y. If we plugged in 5, we would get y = 4. You can put that point in the graph as well, and then draw a parabola that has that vertex and goes through the second point.

Main content

Graphing quadratic functions | Lesson

Google Classroom

What are quadratic functions, and how frequently do they appear on the test?

In a quadratic function, the

of the function is based on an expression in which the

is the highest power term. For example,

f (x) = x^{2} + 2 x + 1

is a quadratic function, because in the highest power term, the

x

is raised to the second power.

Unlike the graphs of linear functions, the graphs of quadratic functions are nonlinear: they don't look like straight lines. Specifically, the graphs of quadratic functions are called parabolas.

On your official SAT, you'll likely see 2 to 4 questions that test your understanding of the connection between quadratic functions and parabolas.

You can learn anything. Let's do this!

How do I graph parabolas, and what are their features?

Parabolas intro

Khan Academy video wrapper

Parabolas introSee video transcript

What are the features of a parabola?

All parabolas have a

y

-intercept, a

, and open either upward or downward.

Since the vertex is the point at which a parabola changes from increasing to decreasing or vice versa, it is also either the maximum or minimum $y$ ‍-value of the parabola.

If the parabola opens upward, then the vertex is the lowest point on the parabola.
If the parabola opens downward, then the vertex is the highest point on the parabola.

A parabola can also have zero, one, or two $x$ ‍-intercepts.

Note: the terms "zero" and "root" are used interchangeably with "

x

-intercept". They all mean the same thing!

Parabolas also have vertical symmetry along a vertical line that passes through the vertex.

For example, if a parabola has a vertex at

(2, 0)

, then the parabola has the same

y

-values at

x = 1

and

x = 3

, at

x = 0

and

x = 4

, and so on.

To graph a quadratic function:

Evaluate the function at several different values of $x$ ‍.
Plot the input-output pairs as points in the $x y$ ‍-plane.
Sketch a parabola that passes through the points.

Example: Graph

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

in the

x y

-plane.

[Show me!]

Try it!

TRY: match the features of the parabola to their coordinates

The graph of

y = - x^{2} + 4 x + 5

is shown above. Match the features of the graph with their coordinates.

1

How do I identify features of parabolas from quadratic functions?

Forms & features of quadratic functions

Khan Academy video wrapper

Forms & features of quadratic functionsSee video transcript

Standard form, factored form, and vertex form: What forms do quadratic equations take?

For all three forms of quadratic equations, the coefficient of the

x^{2}

-term,

a

, tells us whether the parabola opens upward or downward:

If $a > 0$ ‍, then the parabola opens upward.
If $a < 0$ ‍, then the parabola opens downward.

The magnitude of

a

also describes how steep or shallow the parabola is. Parabolas with larger magnitudes of

a

are more steep and narrow compared to parabolas with smaller magnitudes of

a

, which tend to be more shallow and wide.

The graph below shows the graphs of

y = a x^{2}

for various values of

a

The standard form of a quadratic equation,

y = a x^{2} + b x + c

, shows the

y

-intercept of the parabola:

The $y$ ‍-intercept of the parabola is located at $(0, c)$ ‍.

The factored form of a quadratic equation,

y = a (x - b) (x - c)

, shows the

x

-intercept(s) of the parabola:

$x = b$ ‍ and $x = c$ ‍ are solutions to the equation $a (x - b) (x - c) = 0$ ‍.
The $x$ ‍-intercepts of the parabola are located at $(b, 0)$ ‍ and $(c, 0)$ ‍.
The terms $x$ ‍-intercept, zero, and root can be used interchangeably.

The vertex form of a quadratic equation,

y = a (x - h)^{2} + k

, reveals the vertex of the parabola.

The vertex of the parabola is located at $(h, k)$ ‍.

To identify the features of a parabola from a quadratic equation:

Remember which equation form displays the relevant features as constants or coefficients.
Rewrite the equation in a more helpful form if necessary.
Identify the constants or coefficients that correspond to the features of interest.

Example: What are the zeros of the graph of

f (x) = x^{2} + 7 x + 12

[Show me!]

To match a parabola with its quadratic equation:

Determine the features of the parabola.
Identify the features shown in quadratic equation(s).
Select a quadratic equation with the same features as the parabola.
Plug in a point that is not a feature from Step 2 to calculate the coefficient of the $x^{2}$ ‍-term if necessary.

Example:

What is a possible equation for the parabola shown above?

[Show me!]

Try it!

TRY: determine the feature of a parabola from its equation

The graph of the equation

y = \frac{1}{2} x^{2} - 7

is a parabola that opens

because the coefficient of

x^{2}

TRY: determine the feature of a parabola from its equation

The

of the graph of

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

is located at the point

(- 2, 7)

7

is the

y

-value of the graph.

TRY: determine the feature of a parabola from its equation

y = (x - 2) (x - 4)

is the

form of a quadratic equation. The

of the graph can be identified as constants in the equation.

How do I transform graphs of quadratic functions?

Intro to parabola transformations

Khan Academy video wrapper

Intro to parabola transformationsSee video transcript

Translating, stretching, and reflecting: How does changing the function transform the parabola?

We can use function notation to represent the translation of a graph in the

x y

-plane. If the graph of

y = f (x)

is graphed in the

x y

-plane and

c

is a positive constant:

The graph of $y = f (x - c)$ ‍ is the graph of $f (x)$ ‍ shifted to the right by $c$ ‍ units.
The graph of $y = f (x + c)$ ‍ is the graph of $f (x)$ ‍ shifted to the left by $c$ ‍ units.
The graph of $y = f (x) + c$ ‍ is the graph of $f (x)$ ‍ shifted up by $c$ ‍ units.
The graph of $y = f (x) - c$ ‍ is the graph of $f (x)$ ‍ shifted down by $c$ ‍ units.

The graph below shows the graph of the quadratic function

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

alongside various translations:

The graph of $f (x - 4) = (x - 4)^{2} - 3$ ‍ translates the graph of $4$ ‍ units to the right.
The graph of $f (x + 6) = (x + 6)^{2} - 3$ ‍ translates the graph $6$ ‍ units to the left.
The graph of $f (x) + 5 = x^{2} + 2$ ‍ translates the graph $5$ ‍ units up.
The graph of $f (x) - 3 = x^{2} - 6$ ‍ translates the graph $3$ ‍ units down.

We can also represent stretching and reflecting graphs algebraically. If the graph of

y = f (x)

is graphed in the

x y

-plane and

c

is a positive constant:

The graph of $y = - f (x)$ ‍ is the graph of $f (x)$ ‍ reflected across the $x$ ‍-axis.
The graph of $y = f (- x)$ ‍ is the graph of $f (x)$ ‍ reflected across the $y$ ‍-axis.
The graph of $y = c \cdot f (x)$ ‍ is the graph of $f (x)$ ‍ stretched vertically by a factor of $c$ ‍.

The graph below shows the graph of the quadratic function

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

alongside various transformations:

The graph of $- f (x) = - x^{2} + 2 x + 2$ ‍ is the graph of $f (x)$ ‍ reflected across the $x$ ‍-axis.
The graph of $f (- x) = x^{2} + 2 x - 2$ ‍ is the graph of $f (x)$ ‍ reflected across the $y$ ‍-axis.
The graph of $3 \cdot f (x) = 3 x^{2} - 6 x - 6$ ‍ is the graph of $f (x)$ ‍ stretched vertically by a factor of $3$ ‍.

Try it!

TRY: Shift a parabola

Compared to the graph of

y = x^{2}

, the graph of

y = (x + 4)^{2} + 3

is shifted

4

units

and

3

units

Try: reflect a parabola

The graph of

y = f (x)

is a parabola that opens upward and has a vertex located at

(2, 1)

The graph of

y = f (- x)

has a vertex located at

The graph of

y = - f (x)

is a parabola that opens

Your turn!

Practice: identify the features of a parabola

If the function

f (x) = (x - 3)^{2} - 11

is graphed in the

x y

-plane, what are the coordinates of the vertex?

Practice: match a parabola to a quadratic function

Function

f

is graphed in the

x y

-plane above. Which of the following could be

f

Things to remember

Forms of quadratic equations

Standard form: A parabola with the equation

y = a x^{2} + b x + c

has its

y

-intercept located at

(0, c)

Factored form: A parabola with the equation

y = a (x - b) (x - c)

has its

x

-intercept(s) located at

(b, 0)

and

(c, 0)

Vertex form: A parabola with the equation

y = a (x - h)^{2} + k

has its vertex located at

(h, k)

Transformations

If the graph of

y = f (x)

is graphed in the

x y

-plane and

c

is a positive constant:

The graph of $y = f (x - c)$ ‍ is the graph of $f (x)$ ‍ shifted to the right by $c$ ‍ units.
The graph of $y = f (x + c)$ ‍ is the graph of $f (x)$ ‍ shifted to the left by $c$ ‍ units.
The graph of $y = f (x) + c$ ‍ is the graph of $f (x)$ ‍ shifted up by $c$ ‍ units.
The graph of $y = f (x) - c$ ‍ is the graph of $f (x)$ ‍ shifted down by $c$ ‍ units.
The graph of $y = - f (x)$ ‍ is the graph of $f (x)$ ‍ reflected across the $x$ ‍-axis.
The graph of $y = f (- x)$ ‍ is the graph of $f (x)$ ‍ reflected across the $y$ ‍-axis.
The graph of $y = c \cdot f (x)$ ‍ is the graph of $f (x)$ ‍ stretched vertically by a factor of $c$ ‍.

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Sort by:

IZH1
Posted 4 years ago. Direct link to IZH1's post “Is it possible to find th...”
Is it possible to find the vertex of the parabola using the equation -b/2a as well as the other equations listed in the article?
Button navigates to signup pageButton navigates to signup page
(11 votes)
Answer
- eva saramya
  Posted 3 years ago. Direct link to eva saramya's post “yes, it is possible, you ...”
  yes, it is possible, you will need to use -b/2a for the x coordinate of the vertex and another formula k=c- b^2/4a for the y coordinate of the vertex. good luck, hope this helped
  Comment on eva saramya's post “yes, it is possible, you ...”
  (7 votes)
Gracie Lynch
Posted 2 years ago. Direct link to Gracie Lynch's post “How do you get the formul...”
How do you get the formula from looking at the parabola? I am having trouble when I try to work backward with what he said.
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Hecretary Bird
  Posted 2 years ago. Direct link to Hecretary Bird's post “You can get the formula f...”
  You can get the formula from looking at the graph of a parabola in two ways: Either by considering the roots of the parabola or the vertex. A parabola is not like a straight line that you can find the equation of if you have two points on the graph, because there are multiple different parabolas that can go through a given set of two points. Instead you need three points, or the vertex and a point.
  
  In the last practice problem on this article, you're asked to find the equation of a parabola. Think about how you can find the roots of a quadratic equation by factoring. The same principle applies here, just in reverse. You can figure out the roots (x-intercepts) from the graph, and just put them together as factors to make an equation. In this form, the equation for a parabola would look like y = a(x - m)(x - n). "a" is a coefficient (responsible for vertically stretching/flipping the parabola and thus doesn't affect the roots), and the roots of the graph are at x = m and x = n. Because the graph in the problem has roots at 3 and -1, our equation would look like y = a(x + 1)(x - 3). The only one that fits this is answer choice B), which has "a" be -1.
  
  You can also find the equation of a quadratic equation by finding the coordinates of the vertex from a graph, then plugging that into vertex form, and then picking a point on the parabola to use in order to solve for your "a" value. Thirdly, I guess you could also use three separate points to put in a system of three equations, which would let you solve for the "a", "b", and "c" in the standard form of a quadratic, but that's too much work for the SAT.
  Button navigates to signup page
  (6 votes)
namra guleen
Posted 3 years ago. Direct link to namra guleen's post “my sat is on 13 of march(...”
my sat is on 13 of march(probably after5 days ) n i'm craming over maths I just need 500 to 600 score for math so which topics should I focus on more ??
Button navigates to signup pageComment on namra guleen's post “my sat is on 13 of march(...”
(2 votes)
Answer
- Kyle Theodore
  Posted 3 years ago. Direct link to Kyle Theodore's post “Also, remember not to str...”
  Also, remember not to stress out over it. Make sure to get a full nights. Good luck on your exam!
  Button navigates to signup page
  (5 votes)
Aranea
Posted 2 years ago. Direct link to Aranea's post “Please help me get access...”
Please help me get access to questions from "Graphing Quadratic Functions"
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(4 votes)
Answer
0070193150
Posted 2 years ago. Direct link to 0070193150's post “how would i graph this th...”
how would i graph this though f(x)=2(x-3)^2-2
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Hecretary Bird
  Posted 2 years ago. Direct link to Hecretary Bird's post “Following the steps in th...”
  Following the steps in the article, you would graph this function by following the steps to transform the parent function of y = x^2.
  Here, we see that 3 is subtracted from x inside the parentheses, which means that we translate right by 3. We subtract 2 from the final answer, so we move down by 2. Our vertex will then be right 3 and down 2 from the normal vertex (0,0), at (3, -2).
  From here, we see that there's a coefficient outside the parentheses, which means we vertically stretch the function by a factor of 2.
  The easiest way to graph this would be to find the vertex and direction that it opens, and then plug in a point for x and see what you get for y. If we plugged in 5, we would get y = 4. You can put that point in the graph as well, and then draw a parabola that has that vertex and goes through the second point.
  Button navigates to signup page
  (2 votes)

SAT (Fall 2023)

Course: SAT (Fall 2023) > Unit 6

Graphing quadratic functions | Lesson

What are quadratic functions, and how frequently do they appear on the test?

How do I graph parabolas, and what are their features?

Parabolas intro

What are the features of a parabola?

Try it!

How do I identify features of parabolas from quadratic functions?

Forms & features of quadratic functions

Standard form, factored form, and vertex form: What forms do quadratic equations take?

Try it!

How do I transform graphs of quadratic functions?

Intro to parabola transformations

Translating, stretching, and reflecting: How does changing the function transform the parabola?

Try it!

Your turn!

Things to remember

Forms of quadratic equations

Transformations

Want to join the conversation?