QuestionAugust 26, 2025

1. y=x^2+6x+9 x-intercept(s) __ y-intercept __ y-intercept __ domain __ range __

1. y=x^2+6x+9 x-intercept(s) __ y-intercept __ y-intercept __ domain __ range __
1. y=x^2+6x+9
x-intercept(s) __
y-intercept __
y-intercept __
domain __
range __

Solution
4.0(203 votes)

Answer

x-intercept(s): x = -3 ### y-intercept: y = 9 ### domain: (-\infty, \infty) ### range: [0, \infty) Explanation 1. Find x-intercepts Set y = 0: x^2 + 6x + 9 = 0. Factor: (x+3)^2 = 0. Solve: x = -3. 2. Find y-intercept Set x = 0: y = 0^2 + 6 \cdot 0 + 9 = 9. 3. Determine domain Quadratic functions have domain of all real numbers: (-\infty, \infty). 4. Determine range Vertex form: y = (x+3)^2. Minimum value at vertex y = 0. Range: [0, \infty).

Explanation

1. Find x-intercepts<br /> Set $y = 0$: $x^2 + 6x + 9 = 0$. Factor: $(x+3)^2 = 0$. Solve: $x = -3$.<br /><br />2. Find y-intercept<br /> Set $x = 0$: $y = 0^2 + 6 \cdot 0 + 9 = 9$.<br /><br />3. Determine domain<br /> Quadratic functions have domain of all real numbers: $(-\infty, \infty)$.<br /><br />4. Determine range<br /> Vertex form: $y = (x+3)^2$. Minimum value at vertex $y = 0$. Range: $[0, \infty)$.
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