a million) f(x) = x² and g(x) = a million/x Now f o g(x) = f(g(x)) = f(a million/x) = a million/x² + a million 2) f(x) = (-2/3)x - 12 enable f(x) = 0 0 = (-2/3)x - 12 Multiply each thing be 3: 3(0) = (3)(-2/3)x - 12(3) 0 = -2x - 36 36 = -2x x = -18 as a result, the 0 is at x = -18 or on the component (-18, 0). 3) Write this equation interior the kind of y = mx + b: x + 3y = 9 y = (-a million/3)x + 3 as a result, any equation that's perpendicular to this line might have a slope that's a destructive reciprocal to -a million/3, meaning that it ought to have a slope of m = 3. as a result, any equation perpendicular to the single given ought to look like: y = 3x + b, for some b. desire this permits!
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Verified answer
You should work it out yourself.
step1.To find the line, you first need to find an x and a y point.
you already have x=3, so you can use it to get the y by substituting the x(ie find f(3)).
step2.you can then use the new equation to get the lines gradient(m) by writing it in y = mx + c form.
step3.the gradient(m) of the line perpendicular to the old line is m(new)= -1/m(old)
step4.use the new m with your current point (the x and the y) to get the equation of the line
that's((y-y(from step1) = m(x-x(given as 3)))
a million) f(x) = x² and g(x) = a million/x Now f o g(x) = f(g(x)) = f(a million/x) = a million/x² + a million 2) f(x) = (-2/3)x - 12 enable f(x) = 0 0 = (-2/3)x - 12 Multiply each thing be 3: 3(0) = (3)(-2/3)x - 12(3) 0 = -2x - 36 36 = -2x x = -18 as a result, the 0 is at x = -18 or on the component (-18, 0). 3) Write this equation interior the kind of y = mx + b: x + 3y = 9 y = (-a million/3)x + 3 as a result, any equation that's perpendicular to this line might have a slope that's a destructive reciprocal to -a million/3, meaning that it ought to have a slope of m = 3. as a result, any equation perpendicular to the single given ought to look like: y = 3x + b, for some b. desire this permits!