Given f(x) = x^2 + 1 and g(x) = x-2
a. Find (f-g)(-2)
[f-g](x) = f(x) - g(x) = x^2-x+3
[f-g](-2) = (-2)^2-(-2)+3 = 9
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b. Find f[g(5)]
f[g(5)] = f[5-2] = f[3] = 9+1 = 10
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problem a.
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(f-g)(x) = f(x) - g(x)
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(f-g)(-2) = f(-2) - g(-2)
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f(x) = x^2 + 1
f(-2) = (-2)^2 + 1
f(-2) = 4+1
f(-2) = 5
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g(x) = x-2
g(-2) = -2-2
g(-2) = -4
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f-g(-2) = f(-2) - g(-2) = 5 - (-4) = 5 + 4 = 9
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answer for a is:
f-g(-2) = 9
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problem b.
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g(x) = x-2
g(5) = 5-2
g(5) = 3
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f(x) = x^2 + 1
f(g(5)) = (g(5))^2 + 1
since g(5) = 3, equation becomes:
f(g(5)) = 3^2 + 1
f(g(5)) = 9 + 1 = 10
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answer for b is:
f(g(5)) = 10
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in general, you substitute whatever value is replacing x in the equation to get your answers.
looking at problem b in this way, we would get a general solution as follows:
f(x) = x^2 + 1
g(x) = x-2
substitute g(x) for x:
f(g(x)) = (g(x))^2 + 1
substitute the equation for g(x) on the right hand side.
f(g(x)) = (x-2)^2 + 1
remove parentheses:
f(g(x)) = x^2 - 4*x + 4 + 1
simplify:
f(g(x)) = x^2 - 4*x + 5
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substituting 5 for x:
f(g(5)) = (5^2 - 4*5 + 5
simplifying:
f(g(5)) = 25 - 20 + 5
f(g(5)) = 10
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answer is the same as above where we first solved for g(5) which became 3, and then substituted that value in f(g(x)) which made it f(3)).
Hope this helps!
y = .0775x + 700
y = .0775(4,000) + 700
y = 310 + 700
y = 1,010
danielle earned $1,010 last week if she sold $4,000 in electronics merchandise.
as a decimal is <span>0.385</span>
The answer is m = 250 - 25w.
This is because this answer is in y=mx+b format (slope-intercept form) and has the correct values.
In slope-intercept form, m is the slope and b is y-intercept.
In m = 250 - 25w, the y-intercept is 250 and the slope is -25.
This matches the graph.
Y-intercept is where a line crosses the y-axis, which is 250 in this case.
The slope is negative since the values are decreasing over time (as x-values increase, y-values decrease).
The red graph is the result of a vertical translation, 4 units down, of the black graph.
This means that, whenever the black graph associates
The red graph must associate 4 less to the same input:
So, the equation for the red graph is
or, equivalently,
