Match the Quadratic Function y=-(x-4)² to Its Graph: Visual Analysis

Q: Why is the vertex at (4,0) and not (-4,0)?

In vertex form $ y = a(x-h)^2 + k $, the vertex is at (h,k). When you see $ (x-4)^2 $, that means h = 4, so the vertex x-coordinate is positive 4!

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

Look at the coefficient a in front of the squared term. If a is negative (like -1 in this problem), the parabola opens downward. If a is positive, it opens upward.

Q: What if there's no +k term at the end?

When there's no constant term added, it means k = 0, so the vertex sits right on the x-axis. That's why our vertex is at (4,0) instead of (4, some other number).

Q: How can I double-check I picked the right graph?

Test a point! Pick an easy x-value like x = 5. Then $ y = -(5-4)^2 = -(1)^2 = -1 $. The point (5,-1) should be on your chosen graph.

Q: Why does the parabola look symmetric?

All parabolas are symmetric around their vertex. The line of symmetry is $ x = 4 $ in this case, meaning points equidistant from x = 4 have the same y-value.

Vertex Form Parabolas with Downward Opening

One function

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

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 negative, meaning a sad parabola

00:07 The term P equals (4)

00:12 The term K equals (0)

00:15 X-axis intersection points according to the terms

00:23 Let's draw the function according to the intersection points and parabola type

00:30 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

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

for the corresponding chart

Step-by-step solution

The problem involves matching a given function $y = -(x-4)^2$ with its corresponding graph from multiple choices.

First, let's analyze the function:

The function is in vertex form $y = a(x-h)^2 + k$ , where the vertex is $(h, k)$ .
Here, $a = -1$ , $h = 4$ , and $k = 0$ , so the vertex is $(4, 0)$ .
The negative sign indicates the parabola opens downward.

To match the function with the correct graph:

Identify that the vertex of the parabola is at $(4, 0)$ .
The parabola is symmetric around the line $x = 4$ , opening downwards.

Upon examining the choices, Option 1 clearly shows a parabola with a vertex at $(4, 0)$ opening downward. This matches perfectly with the function $y = -(x-4)^2$ .

Therefore, the correct answer to the problem is 1.

Final Answer

Key Points to Remember

Essential concepts to master this topic

Vertex Form: $y = a(x-h)^2 + k$ shows vertex at (h,k)
Direction: Negative coefficient a = -1 means parabola opens downward
Verification: Check vertex (4,0) and downward opening on correct graph ✓

Common Mistakes

Avoid these frequent errors

Confusing the signs in vertex form
Don't think the vertex is at (-4,0) when you see (x-4) = wrong vertex location! The minus sign in (x-h) means the vertex x-coordinate is positive h. Always remember: $y = -(x-4)^2$ has vertex at (4,0), not (-4,0).

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 (4,0) and not (-4,0)?

In vertex form $y = a(x-h)^2 + k$ , the vertex is at (h,k). When you see $(x-4)^2$ , that means h = 4, so the vertex x-coordinate is positive 4!

How do I know which way the parabola opens?

Look at the coefficient a in front of the squared term. If a is negative (like -1 in this problem), the parabola opens downward. If a is positive, it opens upward.

What if there's no +k term at the end?

When there's no constant term added, it means k = 0, so the vertex sits right on the x-axis. That's why our vertex is at (4,0) instead of (4, some other number).

How can I double-check I picked the right graph?

Test a point! Pick an easy x-value like x = 5. Then $y = -(5-4)^2 = -(1)^2 = -1$ . The point (5,-1) should be on your chosen graph.

Why does the parabola look symmetric?

All parabolas are symmetric around their vertex. The line of symmetry is $x = 4$ in this case, meaning points equidistant from x = 4 have the same y-value.

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