Solving y=(x+2)²-4: Finding X-Axis Intersections

Q: What does it mean to find where a function intersects the x-axis?

X-intercepts are points where the graph crosses or touches the x-axis. At these points, the y-coordinate is always zero, so we set the function equal to 0 and solve for x.

Q: Why do I get two answers when solving $ (x+2)^2 = 4 $ ?

Because two different numbers can square to give 4: both 2 and -2. So $ x+2 $ can equal either +2 or -2, giving us two solutions for x.

Q: What if I expanded $ (x+2)^2 $ first instead?

That works too! You'd get $ x^2 + 4x + 4 - 4 = 0 $, which simplifies to $ x^2 + 4x = 0 $. Factor: $ x(x+4) = 0 $ gives the same answers: x = 0 and x = -4.

Q: Do all parabolas have exactly two x-intercepts?

Not always! A parabola can have two x-intercepts (like this one), one x-intercept (when it just touches the x-axis), or zero x-intercepts (when it never touches the x-axis).

Quadratic Functions with X-Intercept Solutions

Find the intersection of the function

$y=(x+2)^2-4$

With the X

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the intersection point with the X-axis

00:04 Substitute Y=0 and solve to find the intersection point

00:12 We want to isolate X

00:19 Extract the root

00:28 When extracting a root there are 2 solutions, positive and negative

00:35 Solve each possibility to find the intersection points

01:05 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the intersection of the function

$y=(x+2)^2-4$

With the X

Step-by-step solution

To solve this problem, we need to determine where the parabola $y = (x+2)^2 - 4$ intersects the x-axis. This occurs where $y = 0$ .

Step 1: Set the equation equal to zero to find the x-intercepts:
$0 = (x+2)^2 - 4$

Step 2: Simplify the equation:
$(x+2)^2 = 4$

Step 3: Solve for $x$ by taking the square root of both sides:
$x+2 = \pm 2$

Step 4: Solve each equation for $x$ :
1. $x+2 = 2$ leads to $x = 0$
2. $x+2 = -2$ leads to $x = -4$

Therefore, the points of intersection are $(-4, 0)$ and $(0, 0)$ , where the parabola intersects the x-axis.

The correct answer to the problem is $(-4, 0), (0, 0)$ .

Final Answer

$(-4,0),(0,0)$

Key Points to Remember

Essential concepts to master this topic

X-Intercepts: Set function equal to zero to find where graph crosses x-axis
Square Root Method: From $(x+2)^2 = 4$ , take $\pm \sqrt{4} = \pm 2$
Verification: Check both solutions: $(-4+2)^2-4 = 0$ and $(0+2)^2-4 = 0$ ✓

Common Mistakes

Avoid these frequent errors

Forgetting the ± when taking square roots
Don't solve $(x+2)^2 = 4$ as just $x+2 = 2$ = only one solution! This misses half the answer because squaring eliminates sign information. Always use $x+2 = \pm 2$ to get both x-intercepts.

Practice Quiz

Test your knowledge with interactive questions

Which equation represents the function:

$$ y=x^2 $$

moved 2 spaces to the right

and 5 spaces upwards.

FAQ

Everything you need to know about this question

What does it mean to find where a function intersects the x-axis?

X-intercepts are points where the graph crosses or touches the x-axis. At these points, the y-coordinate is always zero, so we set the function equal to 0 and solve for x.

Why do I get two answers when solving $(x+2)^2 = 4$ ?

Because two different numbers can square to give 4: both 2 and -2. So $x+2$ can equal either +2 or -2, giving us two solutions for x.

How can I check if (-4,0) and (0,0) are really on the graph?

Substitute each x-value back into the original function. For x = -4: $y = (-4+2)^2 - 4 = 4 - 4 = 0$ ✓. For x = 0: $y = (0+2)^2 - 4 = 4 - 4 = 0$ ✓

What if I expanded $(x+2)^2$ first instead?

That works too! You'd get $x^2 + 4x + 4 - 4 = 0$ , which simplifies to $x^2 + 4x = 0$ . Factor: $x(x+4) = 0$ gives the same answers: x = 0 and x = -4.

Do all parabolas have exactly two x-intercepts?

Not always! A parabola can have two x-intercepts (like this one), one x-intercept (when it just touches the x-axis), or zero x-intercepts (when it never touches the x-axis).

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Solving y=(x+2)²-4: Finding X-Axis Intersections

Quadratic Functions with X-Intercept Solutions

❤️ Continue Your Math Journey!

Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

What does it mean to find where a function intersects the x-axis?

Why do I get two answers when solving $(x+2)^2 = 4$ ?

How can I check if (-4,0) and (0,0) are really on the graph?

What if I expanded $(x+2)^2$ first instead?

Do all parabolas have exactly two x-intercepts?

🌟 Unlock Your Math Potential

More Questions

Parabola of the Form y=(x-p)²+k

Find X-Intercepts of y=(x+1)²-5: Quadratic Function Analysis

Find X-Axis Intersections of y=(x+3)²+4: Quadratic Function Analysis

Solving y=(x+2)²-4: Finding X-Axis Intersections

Quadratic Functions with X-Intercept Solutions

❤️ Continue Your Math Journey!

Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

What does it mean to find where a function intersects the x-axis?

Why do I get two answers when solving (x+2)2=4 (x+2)^2 = 4 (x+2)2=4?

How can I check if (-4,0) and (0,0) are really on the graph?

What if I expanded (x+2)2 (x+2)^2 (x+2)2 first instead?

Do all parabolas have exactly two x-intercepts?

🌟 Unlock Your Math Potential

More Questions

Parabola of the Form y=(x-p)²+k

Find X-Intercepts of y=(x+1)²-5: Quadratic Function Analysis

Find X-Axis Intersections of y=(x+3)²+4: Quadratic Function Analysis

Why do I get two answers when solving $(x+2)^2 = 4$ ?

What if I expanded $(x+2)^2$ first instead?