Vertex Form Practice Problems - Quadratic Functions

Master vertex form Y=a(X-p)²+c with step-by-step practice problems. Find vertex coordinates, convert equations, and solve quadratic functions easily.

📚Practice Vertex Form of Quadratic Functions

Identify vertex coordinates (p,c) from vertex form equations Y=a(X-p)²+c
Convert quadratic functions from standard form to vertex form step-by-step
Determine parabola direction and opening width using coefficient 'a'
Solve real-world problems involving vertex form of quadratic equations
Graph parabolas using vertex coordinates and transformation rules
Master the minus sign rule in vertex form template Y=a(X-p)²+c

Understanding Vertex Representation

Complete explanation with examples

Vertex form of the quadratic equation

The vertex form allows us to identify, very easily, the vertex of the parabola and hence its name.

The vertex form of the quadratic function is:
$Y=a(X-p)^2+c$

Detailed explanation

Practice Vertex Representation

Test your knowledge with 26 quizzes

Find the standard representation of the following function

$$ f(x)=(x+4)^2-16 $$

Examples with solutions for Vertex Representation

Step-by-step solutions included

Exercise #1

Find the standard representation of the following function:

$f(x)=(x-3)^2+x$

Step-by-Step Solution

To solve this problem, we'll perform these steps:

Expand $(x-3)^2$ using the formula for a square:
$(x-3)^2 = x^2 - 2 \times x \times 3 + 3^2 = x^2 - 6x + 9$
Add the $x$ from $f(x) = (x-3)^2 + x$ to the expanded terms:
$x^2 - 6x + 9 + x$
Combine like terms:
$x^2 - 6x + x + 9 = x^2 - 5x + 9$

Therefore, the standard form of the function $f(x)$ is $f(x) = x^2 - 5x + 9$ .

Thus, the correct choice is Choice 3.

Answer:

$f(x)=x^2-5x+9$

Video Solution

Exercise #2

Find the vertex of the parabola

$y=(x+1)^2$

Step-by-Step Solution

The equation $y = (x+1)^2$ is already in the vertex form $y = (x-h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.

By comparing, we have:

The expression inside the square is $(x+1)$ , which can be rewritten as $(x - (-1))$ . Thus, $h = -1$ .
The term $k$ is not present, which means $k = 0$ .

Therefore, the vertex $(h, k)$ of the parabola is $(-1, 0)$ .

Thus, the correct answer is $(-1, 0)$ .

Answer:

$(-1,0)$

Video Solution

Exercise #3

Find the vertex of the parabola

$y=(x-1)^2-1$

Step-by-Step Solution

The given equation is $y = (x-1)^2 - 1$ . This equation is in the vertex form, $y = a(x-h)^2 + k$ , where $a$ , $h$ , and $k$ are constants.

In this case, the given equation can be written as $y = 1 \cdot (x-1)^2 + (-1)$ , indicating that $a = 1$ , $h = 1$ , and $k = -1$ .

The vertex form of the quadratic equation allows us to directly identify the vertex of the parabola as $(h, k)$ .

From our identification, it is clear that the vertex $(h, k)$ of the parabola is $(1, -1)$ .

Therefore, the vertex of the given parabola is $(1, -1)$ .

Answer:

$(1,-1)$

Video Solution

Exercise #4

Find the vertex of the parabola

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

Step-by-Step Solution

The given equation of the parabola is $y = (x+1)^2 - 1$ .

This equation is already in the vertex form, $y = a(x-h)^2 + k$ , where $(h, k)$ is the vertex.

By comparing, we identify:
The expression $(x + 1)$ implies that $h = -1$ (since $x + 1$ is equivalent to $(x - (-1))$ ).
The constant $-1$ is the $k$ value.

Thus, the vertex $(h, k)$ is $(-1, -1)$ .

Therefore, the vertex of the parabola is at the point $(-1,-1)$ .

Answer:

$(-1,-1)$

Video Solution

Exercise #5

Find the standard representation of the following function

$f(x)=(2x+1)^2-1$

Step-by-Step Solution

To convert $f(x) = (2x+1)^2 - 1$ into its standard quadratic form, we need to expand $(2x+1)^2$ first and then adjust for the subtraction of 1.

The expansion is carried out using the binomial expansion formula:

$(2x + 1)^2 = (2x)^2 + 2(2x)(1) + 1^2$ .

Calculating each term gives:

$(2x)^2 = 4x^2$
$2(2x)(1) = 4x$
$1^2 = 1$

Combining these, we obtain:

$(2x + 1)^2 = 4x^2 + 4x + 1$

Now, substituting back into the original equation:

$f(x) = (2x+1)^2 - 1 = (4x^2 + 4x + 1) - 1$

Subtracting 1 from the constant term, we get:

$f(x) = 4x^2 + 4x + 1 - 1 = 4x^2 + 4x$

Therefore, the standard form representation of the function is $f(x) = 4x^2 + 4x$ .

Answer:

$f(x)=4x^2+4x$

Video Solution

Frequently Asked Questions

Everything you need to know about Vertex Representation

What is vertex form of a quadratic function?

Vertex form is Y=a(X-p)²+c where (p,c) represents the vertex coordinates of the parabola. The parameter 'a' determines the parabola's direction and width, while p is the x-coordinate and c is the y-coordinate of the vertex.

How do you find the vertex from vertex form Y=a(X-p)²+c?

The vertex coordinates are (p,c) directly from the equation. Remember there's a minus sign before p in the formula, so if you see Y=2(X-3)²+5, the vertex is (3,5), not (-3,5).

Why is there a minus sign in vertex form Y=a(X-p)²+c?

The minus sign is part of the vertex form template structure. It doesn't mean p is negative. If the actual vertex has a negative x-coordinate, you'll place the negative value in place of p, making the expression positive inside the parentheses.

What does the coefficient 'a' tell you in vertex form?

The coefficient 'a' determines two things: 1) Direction - if a>0, parabola opens upward; if a<0, opens downward. 2) Width - larger |a| values make narrower parabolas, smaller |a| values make wider parabolas.

How do you convert from standard form to vertex form?

Use completing the square method: 1) Factor out 'a' from x² and x terms, 2) Complete the square inside parentheses, 3) Simplify to get Y=a(X-p)²+c format. This process reveals the vertex coordinates directly.

What are common mistakes when working with vertex form?

Common errors include: • Forgetting the minus sign rule (thinking vertex is (-p,c) instead of (p,c)) • Mixing up which coordinate is which (p is x-coordinate, c is y-coordinate) • Incorrectly identifying the coefficient 'a' when converting between forms

When is vertex form most useful for quadratic functions?

Vertex form is most helpful when you need to: quickly identify the vertex for graphing, understand parabola transformations, solve optimization problems, or determine maximum/minimum values in real-world applications.

Can you have fractions or decimals in vertex form?

Yes, vertex form can contain fractions, decimals, or any real numbers. For example, Y=0.5(X-2.5)²+1.25 is valid vertex form with vertex at (2.5, 1.25) and parabola opening upward with width factor 0.5.

Continue Your Math Journey

Master Vertex Representation first, then advance to these related topics that build on your fraction skills

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Vertex Form Practice Problems - Quadratic Functions

Understanding Vertex Representation

Vertex form of the quadratic equation

Practice Vertex Representation

Examples with solutions for Vertex Representation

Frequently Asked Questions

What is vertex form of a quadratic function?

How do you find the vertex from vertex form Y=a(X-p)²+c?

Why is there a minus sign in vertex form Y=a(X-p)²+c?

What does the coefficient 'a' tell you in vertex form?

How do you convert from standard form to vertex form?

What are common mistakes when working with vertex form?

When is vertex form most useful for quadratic functions?

Can you have fractions or decimals in vertex form?

More Vertex Representation Questions

Vertex Representation

Continue Your Math Journey

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Topics Learned in Later Sections

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