The x and y-intercept(s) of the graph of the function f(x)=−4|x−3|+12 are: ( 0, 0 ), ( 6, 0 ) and ( 0, 0 ) respectively.
Given the function;
f(x) = −4| x-3 | + 12
First, we find the x-intercepts by simply substituting 0 for y and solve for x.
f(x) = −4| x-3 | + 12
y = −4| x-3 | + 12
0 = −4| x-3 | + 12
−4| x-3 | + 12 = 0
−4| x-3 | = -12
Divide both sides by -4
| x-3 | = 3
We remove the absolute value sign
x - 3 = ±3
x = 3±3
x = 3-3, 3+3
x = 0, 3
Hence, x-intercepts are: ( 0, 0 ), ( 6, 0 )
Next, we find the y-intercept by substituting 0 for x and solve for y.
f(x) = −4| x-3 | + 12
y = −4| 0-3 | + 12
y = −4| -3 | + 12
We remove the absolute value sign
y = −4(3) + 12
y = −12 + 12
y = 0
Hence, y-intercept is: ( 0, 0 ).
The x and y-intercept(s) of the graph of the function f(x)=−4|x−3|+12 are: ( 0, 0 ), ( 6, 0 ) and ( 0, 0 ) respectively.
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