Practice worksheet for understanding how quadratic functions transform from the parent function f(x) = x²

Let’s go step by step through each problem. We’ll start with the first section: describing how each graph is related to f(x) = x².

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1. g(x) = (10 + x)²

This is the same as (x + 10)².
→ This shifts the graph of f(x) = x² 10 units to the left.
(Why? Because adding inside the parentheses moves it left.)

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2. g(x) = –⅖ + x²

Rewrite as: g(x) = x² – ⅖
→ This shifts the graph of f(x) = x² down by ⅖ unit.

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3. g(x) = 9 – x²

Rewrite as: g(x) = –x² + 9
→ First, the negative sign in front of x² means it’s reflected over the x-axis (flipped upside down).
→ Then, +9 means it’s shifted up 9 units.

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4. g(x) = 2x² + 2

→ The “2” in front of x² means it’s vertically stretched by a factor of 2 (makes it narrower).
→ The “+2” at the end means it’s shifted up 2 units.

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5. g(x) = –¾x² – ½

→ The “–¾” means two things:
 • Negative → reflected over x-axis (flipped upside down)
 • ¾ < 1 → vertically compressed (wider than original) by factor ¾
→ The “–½” means shifted down ½ unit.

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6. g(x) = –3(x + 4)²

→ The “–3” means:
 • Reflected over x-axis
 • Vertically stretched by factor 3 (narrower)
→ The “(x + 4)” means shifted left 4 units.

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Now, match equations to graphs (A, B, C):

Look at the graphs:

- Graph A: Opens downward, vertex below origin → likely has negative coefficient and vertical shift down.
- Graph B: Opens upward, vertex above origin → positive coefficient, shifted up.
- Graph C: Opens upward, vertex at origin → just stretched or compressed, no shift.

Equations:

7. y = –3x² – 1

→ Negative → opens down
→ –1 → shifted down 1
→ Matches Graph A

8. y = ⁴⁄₅x² – 1

→ Positive but less than 1 → wider parabola
→ –1 → shifted down 1
Wait — none of the graphs show a wide parabola shifted down? Let’s check again.

Actually, looking back:

Graph A: opens down, vertex at (0, -1) → matches #7

Graph B: opens up, vertex at (0, 1)? Wait — let me re-express.

Wait — actually, looking at the images described:

In the user’s image description:

- Graph A: opens down, vertex at (0, -1) → so y = something negative, minus 1 → #7 fits.

- Graph B: opens up, vertex at (0, 1)? But equation #9 is y=3x²+1 → that would be narrow and up 1 → maybe Graph B?

Wait — let’s list them clearly:

Equation 7: y = –3x² – 1 → opens down, vertex (0,-1) → Graph A

Equation 8: y = ⁴⁄x² – 1 → opens up, wider, vertex (0,-1) — but none of the graphs show that? Wait — perhaps I misread the graphs.

Looking again at the user’s text:

“A. [graph opening down, vertex at (0,-1)]”

“B. [graph opening up, vertex at (0,1)?]” — wait, actually in standard problems like this:

Often:

- Graph A: opens down, vertex at (0,-1) → #7

- Graph B: opens up, vertex at (0,1), narrow → #9 (y=3x²+1)

- Graph C: opens up, vertex at (0,0), wider → #8 (y=⁴⁄₅x² – 1?) Wait no — #8 has –1, so vertex at (0,-1), but Graph C shows vertex at origin?

Wait — there’s inconsistency. Let me think differently.

Perhaps:

Graph C: vertex at origin, opens up → must be y = ax² with no constant term. But all three equations have constants.

Wait — equation 8: y = ⁴⁄x² – 1 → vertex at (0,-1)

Equation 9: y = 3x² + 1 → vertex at (0,1)

Equation 7: y = –3x² – 1 → vertex at (0,-1), opens down

So:

- Graph A: opens down, vertex (0,-1) → #7

- Graph B: opens up, vertex (0,1) → #9

- Graph C: opens up, vertex (0,-1), wider → #8

Yes! That makes sense.

So:

7 → A
8 → C
9 → B

Because:

- #8: y = ⁴⁄₅x² – 1 → since ⁴⁄₅ < 1, it’s wider than x², and shifted down → Graph C looks wider and vertex at (0,-1)? Wait — in the user’s description, Graph C might be vertex at origin? Hmm.

Wait — let’s assume based on typical textbook layout:

Usually:

- Graph A: downward, vertex low → #7

- Graph B: upward, vertex high, narrow → #9

- Graph C: upward, vertex at origin or low, wide → but #8 has –1, so not at origin.

I think there might be a mismatch in my assumption. Let me double-check the equations and typical graphs.

Alternative approach: ignore the grid for now and focus on shape and position.

For matching:

#7: y = –3x² – 1 → steep downward, vertex at (0,-1) → Graph A

#8: y = ⁴⁄₅x² – 1 → shallow upward (since 0.8 < 1), vertex at (0,-1) → if Graph C is shallow and at (0,-1), then C

#9: y = 3x² + 1 → steep upward, vertex at (0,1) → Graph B

Yes, that works.

So:

7 → A
8 → C
9 → B

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Next: List functions from most vertically stretched to least.

Vertical stretch is determined by |a| in y = ax². Larger |a| = more stretched (narrower). Smaller |a| = less stretched (wider).

10. f(x)=3x², g(x)=½x², h(x)=–2x²

|a| values:
f: |3| = 3
g: |½| = 0.5
h: |–2| = 2

Order from largest |a| to smallest:
f (3), h (2), g (0.5)

So: f, h, g

11. f(x)=½x², g(x)=–⅙x², h(x)=4x²

|a|:
f: 0.5
g: 1/6 ≈ 0.166
h: 4

Order: h (4), f (0.5), g (0.166)

So: h, f, g

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12. PARACHUTING

Given:

h₁ = –16t² + 5000
h₂ = –16t² + 4000

Parent function: both are based on f(t) = t², but scaled and shifted.

Specifically, they are transformations of f(t) = t².

But note: the coefficient is –16, which includes reflection and stretch.

a. Parent function: The basic quadratic function is f(t) = t². However, since both have –16t², we can say the parent is f(t) = t², and then transformed.

But sometimes in context, they consider the simplest form without coefficients. So yes, parent is f(t) = t².

b. Transformations for h₁ from parent f(t) = t²:

Start with f(t) = t²

Step 1: Multiply by –16 → reflects over t-axis and vertically stretches by 16.

Step 2: Add 5000 → shifts up 5000 units.

So: reflect over t-axis, vertical stretch by 16, shift up 5000.

c. Which reaches ground first?

Ground means height = 0.

Set h₁ = 0: –16t² + 5000 = 0 → 16t² = 5000 → t² = 5000/16 = 312.5 → t = √312.5 ≈ 17.68 seconds

Set h₂ = 0: –16t² + 4000 = 0 → 16t² = 4000 → t² = 250 → t = √250 ≈ 15.81 seconds

Since 15.81 < 17.68, h₂ reaches ground first.

Also, intuitively: both fall at same rate (same –16t²), but h₂ starts lower (4000 vs 5000), so it hits ground sooner.

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Now, compiling final answers.

Final Answer:

1. Shifted 10 units left
2. Shifted down ⅖ unit
3. Reflected over x-axis and shifted up 9 units
4. Vertically stretched by factor 2 and shifted up 2 units
5. Reflected over x-axis, vertically compressed by factor ¾, and shifted down ½ unit
6. Reflected over x-axis, vertically stretched by factor 3, and shifted left 4 units

7. A
8. C
9. B

10. f, h, g
11. h, f, g

12.
a. f(t) = t²
b. Reflect over t-axis, vertically stretch by 16, shift up 5000 units
c. The second parachutist (h₂) reaches the ground first