Solve y=-(x-3)²-1: Vertical Shift of 5 Units

Q: Why do I add to the constant term and not somewhere else?

The constant term controls the vertical position of the entire graph. Adding to it moves every point up by that amount, while changing other parts affects the parabola's shape or horizontal position.

Q: How can I tell if the shift is up or down?

Up: Add a positive number to the functionDown: Subtract a positive number (or add a negative). The question says 'move up 5 spaces,' so we add +5.

Q: Does the vertex form help me see transformations?

Yes! In $ y = a(x-h)^2 + k $, the vertex is at (h, k). When k changes from -1 to +4, the vertex moves from (3, -1) to (3, 4).

Q: What if the problem asked to move down instead?

To move down 5 units, you'd subtract 5: $ y = -(x-3)^2 - 1 - 5 = -(x-3)^2 - 6 $. The vertex would be at (3, -6).

Q: Why doesn't the x-part of the vertex change?

Vertical shifts only affect the y-coordinates of all points. The parabola slides straight up or down, so horizontal positions (x-coordinates) stay the same.

Quadratic Transformations with Vertical Shifts

Which equation represents the the function

$y=-(x-3)^2-1$

moved 5 spaces up?

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Express the new function

00:03 We'll use the formula to shift the function

00:11 Want to shift 5 units horizontally upward, so we'll increase K

00:24 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

Which equation represents the the function

$y=-(x-3)^2-1$

moved 5 spaces up?

Step-by-step solution

To solve this problem, we will follow these steps:

Step 1: Identify the original function.
Step 2: Apply the vertical shift of 5 units upward.
Step 3: Write the resulting equation.

Let's go through each step:

Step 1: The given function is $y = -(x-3)^2 - 1$ . This can be identified as a downward-facing parabola with its vertex at the point $(3, -1)$ .

Step 2: To move the entire function 5 spaces up, we add 5 to the constant term $-1$ in the equation. The effect of this transformation is that the new vertex becomes $(3, -1 + 5) = (3, 4)$ .

Step 3: Updating the function, we have:

$y = -(x-3)^2 - 1 + 5$

Simplify by combining the constants:

$y = -(x-3)^2 + 4$

This transformation results in the function moving 5 units up along the vertical axis to a new equation. The final equation is $y = -(x-3)^2 + 4$ .

Therefore, the solution to the problem is $y=-(x-3)^2+4$ , which is choice 4 from the given options.

Final Answer

$y=-(x-3)^2+4$

Key Points to Remember

Essential concepts to master this topic

Vertical Shift Rule: Add or subtract constant to entire function
Technique: $y = -(x-3)^2 - 1 + 5 = -(x-3)^2 + 4$
Check: Vertex moves from (3, -1) to (3, 4) for 5-unit upward shift ✓

Common Mistakes

Avoid these frequent errors

Adding the shift to the wrong part of the equation
Don't add 5 inside the parentheses like (x-3+5)² = wrong function! This changes horizontal position, not vertical. Always add or subtract the shift amount to the constant term outside all parentheses.

Practice Quiz

Test your knowledge with interactive questions

Find the corresponding algebraic representation of the drawing:

FAQ

Everything you need to know about this question

Why do I add to the constant term and not somewhere else?

The constant term controls the vertical position of the entire graph. Adding to it moves every point up by that amount, while changing other parts affects the parabola's shape or horizontal position.

How can I tell if the shift is up or down?

Up: Add a positive number to the function
Down: Subtract a positive number (or add a negative). The question says 'move up 5 spaces,' so we add +5.

Does the vertex form help me see transformations?

Yes! In $y = a(x-h)^2 + k$ , the vertex is at (h, k). When k changes from -1 to +4, the vertex moves from (3, -1) to (3, 4).

What if the problem asked to move down instead?

To move down 5 units, you'd subtract 5: $y = -(x-3)^2 - 1 - 5 = -(x-3)^2 - 6$ . The vertex would be at (3, -6).

Why doesn't the x-part of the vertex change?

Vertical shifts only affect the y-coordinates of all points. The parabola slides straight up or down, so horizontal positions (x-coordinates) stay the same.

How do I verify my answer is correct?

Check that the new vertex is 5 units higher: original vertex (3, -1) plus 5 gives (3, 4). Also verify that $-(3-3)^2 + 4 = 4$ ✓

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