Consider the function f(x)=x^2+3x-18. Let g be a translation of f so that g has zeros x=-3 and x=-17. Write a rule for g.

Answer 1

f(x) = (x+6)(x-3) which has zeroes 3 and -6

f(x) = x^2 +3x -18

g(x) = (x+17)(x+3) which has zeroes -3 and -17

g(x) = x^2 +20x + 51

the graphs have the f(x) crossing the x axis at 3 and -6, 9 apart

and g(x) crossing the x-axis at -3 and -17 which are 14 apart

f(x) has y intercept -18

g(x) has y intercept 51

complete the squares

f(x) =x^2 + 3x + 9/4 - 6 - 9/4 = (x+3/2)^2 - 33/4

the vertex is (-3/2, - 33/4) = (-1.5, -8.25)

g(x) =x^2 +20x +100 +51 -100 = (x+10)^2 -49

the vertex is (-10,-49)

g(x) is f(x) shifted down by 40 3/4, and left by 8 1/2

40 3/4 = the difference in y values of the vertices -8.5 - - 49

8 1/2 = the difference in x values of the vertices: -1.5 - - 10

the general parabola equation is ax^2 + bx + c where a=1 for both f and g

x^2 + bx + c = (x-b/2)^ + c-b^2/4

with (b/2, c-b^2/4) as the vertex

for f(x) b/2=-3/2 or b=3, and c-b^2/4 =-6, or c=6+9/4 = 33/4 = 8.25

for g(x) b/2 =10 or b=20 and c-b^2/4 = -49 or c =49+400/4 = 149

g(x) = f(x) +7x -43 = f(x)+ (10-3)x + (-49+6) where 10 is the b value of g(x) and 3 is the b value of f(x)

and -6 is the c value of f(x) and -49 is the c value of g(x), call them bg, bf, cg and cf

g(x) = f(x) +(bg-bf)x + (cg-cf). Just subtract the b values and subtract the c values to get the new coefficient of the x term and the new constant term

the transformation is just 2 shifts, left and down the general shape of the parabola is the same for g(x) and f(x) since the coefficient of the x squared term was the same for both