Question

question 2 of 10
use the zeros and the labeled point to write the quadratic function represented by the graph.

a. $y = x^2 - 6x + 8$
b. $y = x^2 + 6x + 8$
c. $y = 2x^2 + 12x + 16$
d. $y = 2x^2 - 4x - 16$

Explanation:

Step1: Identify the zeros

From the graph, the zeros (x-intercepts) are at \( x = 2 \) and \( x = 4 \) (assuming the points are at (2,0) and (4,0) from the graph's x-axis). So the quadratic can be written in factored form as \( y = a(x - 2)(x - 4) \).

Step2: Use the labeled point (6,8)

Substitute \( x = 6 \) and \( y = 8 \) into the factored form:
\( 8 = a(6 - 2)(6 - 4) \)
\( 8 = a(4)(2) \)
\( 8 = 8a \)
Divide both sides by 8: \( a = 1 \)? Wait, no, wait. Wait, maybe I misread the zeros. Wait, looking at the graph again, maybe the zeros are at \( x = -2 \) and \( x = -4 \)? No, the x-axis has 5, so maybe the zeros are at \( x = 2 \) and \( x = 4 \)? Wait, no, the graph is on the right side? Wait, no, the x-axis: let's check the options. Let's test the point (6,8) in each option.

Option A: \( y = x^2 - 6x + 8 \). At \( x = 6 \), \( y = 36 - 36 + 8 = 8 \). Wait, that works? But wait, let's check the zeros. For \( y = x^2 - 6x + 8 \), factor: \( (x - 2)(x - 4) \), zeros at 2 and 4. But the graph's zeros: looking at the graph, the parabola opens upwards, and the zeros are between 0 and 5? Wait, the labeled point is (6,8), so x=6, y=8. Let's check Option D: \( y = 2x^2 - 4x - 16 \). At x=6: \( 2(36) - 24 - 16 = 72 - 40 = 32
eq 8 \). Option C: \( 2x^2 + 12x + 16 \). At x=6: \( 2(36) + 72 + 16 = 72 + 72 + 16 = 160
eq 8 \). Option B: \( x^2 + 6x + 8 \). At x=6: \( 36 + 36 + 8 = 80
eq 8 \). Option A: \( x^2 - 6x + 8 \). At x=6: 36 - 36 + 8 = 8. And zeros at x=2 and x=4 (since \( x^2 - 6x + 8 = (x - 2)(x - 4) \)), which matches the graph's zeros (the two red points on the x-axis, probably at 2 and 4). Wait, but the graph's y-intercept: for Option A, when x=0, y=8. But the graph's y-intercept is at 10? Wait, no, maybe I misread the graph. Wait, the graph's y-axis has 10, and the parabola starts at y=10? No, the graph is a parabola opening upwards, with zeros at, say, 2 and 4, and passing through (6,8). Wait, Option A: y = x² -6x +8. At x=0, y=8. But the graph's y-intercept is at 10? Wait, maybe the graph's y-intercept is 8? Or maybe my initial analysis is wrong. Wait, the problem says "use the zeros and the labeled point". Let's re-express.

Wait, the factored form: if zeros are r and s, then \( y = a(x - r)(x - s) \). Let's assume zeros are 2 and 4 (from Option A's factors). Then \( y = a(x - 2)(x - 4) \). Using (6,8): \( 8 = a(4)(2) \Rightarrow a = 1 \). So \( y = (x - 2)(x - 4) = x² -6x +8 \), which is Option A. But wait, the graph's y-intercept: when x=0, y=8. But the graph shows y=10 at x=0? Wait, maybe the graph is drawn with approximate values. Alternatively, maybe I made a mistake. Wait, let's check Option D: \( y = 2x² -4x -16 \). Factor: 2(x² -2x -8) = 2(x -4)(x +2). Zeros at 4 and -2. At x=6, y=2(36) -24 -16=72-40=32≠8. Option C: 2x² +12x +16=2(x² +6x +8)=2(x+2)(x+4). Zeros at -2 and -4. At x=6, y=2(36+36+8)=2(80)=160≠8. Option B: x² +6x +8=(x+2)(x+4). Zeros at -2 and -4. At x=6, y=36+36+8=80≠8. So only Option A gives y=8 at x=6, and zeros at 2 and 4, which matches the graph's zeros (the two red points on the x-axis, probably at 2 and 4). So the correct answer is A.

Answer:

A. \( y = x^2 - 6x + 8 \)