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

Q: Why are there two x-intercepts for this parabola?

Most parabolas cross the x-axis at two points because they have a U-shape. When you take the square root of both sides, you get $ \pm\sqrt{5} $, giving you two solutions!

Q: What does the vertex form tell me about this parabola?

The vertex form $ y = (x+1)^2 - 5 $ shows the vertex is at (-1, -5). The parabola opens upward and is shifted 1 unit left and 5 units down from the origin.

Q: Why can't I just substitute random x-values to find intercepts?

You could, but it would take forever! Setting y = 0 and solving algebraically gives you the exact x-intercept locations immediately, without guessing.

Q: How do I write the final answer as coordinate pairs?

X-intercepts are points where y = 0. So if your x-values are $ -1+\sqrt{5} $ and $ -1-\sqrt{5} $, write them as $ (-1+\sqrt{5}, 0) $ and $ (-1-\sqrt{5}, 0) $.

X-Intercepts with Vertex Form Parabolas

Find the intersection of the function

$y=(x+1)^2-5$

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

00:13 We want to isolate X

00:26 Extract the root

00:33 When extracting a root there are 2 solutions, positive and negative

00:41 Solve each possibility to find the intersection points

01:08 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

$y=(x+1)^2-5$

With the X

Step-by-step solution

To solve this problem, follow these steps:

The problem asks to find the intersection of the parabola with the x-axis given by the equation $y = (x+1)^2 - 5$ . The x-intercepts occur where $y = 0$ .

Let's solve the equation for $y = 0$ :

$(x+1)^2 - 5 = 0$

Simplify and solve for $x$ :

Add 5 to both sides: $(x+1)^2 = 5$
Take the square root of both sides: $x+1 = \pm \sqrt{5}$
Solve for $x$ by subtracting 1 from both sides: $x = -1 \pm \sqrt{5}$

This gives two solutions:

$x = -1 + \sqrt{5}$
$x = -1 - \sqrt{5}$

Therefore, the x-intercepts are $(-1+\sqrt{5}, 0)$ and $(-1-\sqrt{5}, 0)$ .

The correct answer is then the pair $(-1+\sqrt{5}, 0)$ and $(-1-\sqrt{5}, 0)$ , which matches choice 3:

$(-\sqrt{5}-1,0),(\sqrt{5}-1,0)$

Therefore, the solution to the problem is $(-1+\sqrt{5}, 0)$ and $(-1-\sqrt{5}, 0)$

Final Answer

$(-\sqrt{5}-1,0),(\sqrt{5}-1,0)$

Key Points to Remember

Essential concepts to master this topic

Definition: X-intercepts occur when y = 0 in quadratic functions
Technique: Set $(x+1)^2 - 5 = 0$ then solve $(x+1)^2 = 5$
Check: Substitute both solutions back: $(-1+\sqrt{5}+1)^2 - 5 = 0$ ✓

Common Mistakes

Avoid these frequent errors

Expanding the squared term unnecessarily
Don't expand $(x+1)^2$ to $x^2 + 2x + 1$ = complicated quadratic formula! This creates extra work and more chances for errors. Always work directly with the vertex form by isolating the squared term first.

Practice Quiz

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Find the corresponding algebraic representation of the drawing:

FAQ

Everything you need to know about this question

Why are there two x-intercepts for this parabola?

Most parabolas cross the x-axis at two points because they have a U-shape. When you take the square root of both sides, you get $\pm\sqrt{5}$ , giving you two solutions!

What does the vertex form tell me about this parabola?

The vertex form $y = (x+1)^2 - 5$ shows the vertex is at (-1, -5). The parabola opens upward and is shifted 1 unit left and 5 units down from the origin.

Why can't I just substitute random x-values to find intercepts?

You could, but it would take forever! Setting y = 0 and solving algebraically gives you the exact x-intercept locations immediately, without guessing.

How do I write the final answer as coordinate pairs?

X-intercepts are points where y = 0. So if your x-values are $-1+\sqrt{5}$ and $-1-\sqrt{5}$ , write them as $(-1+\sqrt{5}, 0)$ and $(-1-\sqrt{5}, 0)$ .

Can I approximate the square root values?

For exact answers, keep the radical form. But if you need decimals, $\sqrt{5} \approx 2.236$ , so the intercepts are approximately (-3.236, 0) and (1.236, 0).

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