Identify the Graph of y=(x+1)²: Quadratic Function Matching

Q: Why is the vertex at x = -1 and not x = +1?

In vertex form $ y = (x - h)^2 + k $, the vertex is at (h, k). For $ y = (x + 1)^2 $, rewrite as $ y = (x - (-1))^2 + 0 $, so h = -1 and the vertex is at (-1, 0).

Q: How do I know which direction the parabola opens?

Look at the coefficient of the squared term! Since $ y = (x + 1)^2 $ has a positive coefficient (which is 1), the parabola opens upward like a U-shape.

Q: What if the equation was y = -(x + 1)²?

The negative sign would flip the parabola upside down! The vertex would still be at (-1, 0), but it would open downward and be a maximum point instead of minimum.

Q: How can I check if I picked the right graph?

Test a point! Try x = 0: $ y = (0 + 1)^2 = 1 $. The correct graph should pass through (0, 1) and have its lowest point at (-1, 0).

Q: What's the difference between vertex form and standard form?

Vertex form: $ y = (x - h)^2 + k $ - shows vertex directlyStandard form: $ y = ax^2 + bx + c $ - need to complete the square or use formulas to find vertex

Quadratic Functions with Vertex Identification

One function

$y=(x+1)^2$

for the corresponding chart

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Match the correct graph to the function

00:03 The coefficient of X squared is positive, meaning a smiling parabola

00:08 The term P equals (-1)

00:15 The term K equals (0)

00:19 The X-axis intersection points according to the terms

00:27 Let's draw the function according to the intersection points and type of parabola

00:32 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

One function

$y=(x+1)^2$

for the corresponding chart

Step-by-step solution

The provided function is $y = (x + 1)^2$ , a parabola opening upwards. The vertex of this parabola is at $x = -1$ . We need to find a graph where the turning point (vertex) is at $x = -1$ .

To solve this, closely examine the given graphs to identify the one where the parabola's vertex appears at $x = -1$ on the horizontal axis and reflects a symmetry around this axis indicating a minimum point at $y = 0$ .

Given the choices in the chart:

Choice 1 depicts a different vertex or orientation.
Choice 2 also shows a different shift along the x-axis.
Choice 3 is correctly showing that the vertex occurs at $x = -1$ with the appropriate configuration indicated by the problem’s prediction.
Choice 4 does not match the needed features.

Therefore, the graph that matches the equation $y = (x + 1)^2$ is Choice 3, as it is the only graph with the correct vertex point at $(-1, 0)$ .

Consequently, the solution to this problem is Choice 3.

Final Answer

Key Points to Remember

Essential concepts to master this topic

Vertex Form: For $y = (x - h)^2 + k$ , vertex is at (h, k)
Transform Sign: $y = (x + 1)^2$ means x = -1, so vertex at (-1, 0)
Check Graph: Parabola opens upward with minimum point at x = -1 ✓

Common Mistakes

Avoid these frequent errors

Confusing the sign in vertex form
Don't think $y = (x + 1)^2$ has vertex at x = +1! The plus sign means you subtract 1 from x to get zero inside parentheses. Always remember: (x + 1) = 0 when x = -1, so vertex is at x = -1.

Practice Quiz

Test your knowledge with interactive questions

Find the intersection of the function

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

With the X

FAQ

Everything you need to know about this question

Why is the vertex at x = -1 and not x = +1?

In vertex form $y = (x - h)^2 + k$ , the vertex is at (h, k). For $y = (x + 1)^2$ , rewrite as $y = (x - (-1))^2 + 0$ , so h = -1 and the vertex is at (-1, 0).

How do I know which direction the parabola opens?

Look at the coefficient of the squared term! Since $y = (x + 1)^2$ has a positive coefficient (which is 1), the parabola opens upward like a U-shape.

What if the equation was y = -(x + 1)²?

The negative sign would flip the parabola upside down! The vertex would still be at (-1, 0), but it would open downward and be a maximum point instead of minimum.

How can I check if I picked the right graph?

Test a point! Try x = 0: $y = (0 + 1)^2 = 1$ . The correct graph should pass through (0, 1) and have its lowest point at (-1, 0).

What's the difference between vertex form and standard form?

Vertex form: $y = (x - h)^2 + k$ - shows vertex directly
Standard form: $y = ax^2 + bx + c$ - need to complete the square or use formulas to find vertex

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