Heather is writing a quadratic function that represents a parabola that touches but does not cross the x-axis at x = –6. Which f
unction could Heather be writing? f(x) = x2 + 36x + 12 f(x) = x2 – 36x – 12 f(x) = –x2 + 12x + 36 f(x) = –x2 – 12x – 36
1 answer:
Answer:
f(x) = –x^2 – 12x – 36
Step-by-step explanation:
The parent function, x^2, touches the x-axis at x=0. Translating it 6 units left replaces x with x-(-6) = x+6, so the function is ...
f(x) = (x+6)^2 = x^2 +12x +36
Reflecting the graph across the x-axis doesn't change the x-intercept, so Heather could be writing ...
f(x) = -x^2 -12x -36
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The y values from:
1. ₋6
2. ₋3
3. 0
4. 3
Given the x values and y value as y = 3x
1. when x = ₋2 then y = 3x
y = 3(₋2)
y = ₋6
2. when x = ₋1 then y = 3x
y = 3(₋1)
y = ₋3
3. when x = 0 then y = 3x
y = 3(0)
y = 0
4. when x = 1 then y = 3x
y = 3(1)
y = 3
therefore we get the points as (₋2 , ₋6) (₋1 , ₋3) (0,0) (1 , 3).The graph is pltted and attached.
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