A horizontal stretch is applied to the graph of y = x^3 so that the transformed curve passes through (2, 64). The horizontal stretch factor, rounded to the nearest tenth, is?

Answer 1

64= a(2^3) = 8a

a=64/8 = 4

y=f(x)= 4x^3

or for nearest 1 decimal place

y = f(x) = 4.0x^3

wait. whoops, that's for a vertical stretch

rework the numbers

1st off hand guess

y= (1/3)x^3, the inverse coefficient of the vertical stretch

but maybe not

how about y=(1/3)x^2+ 61 1/3? but that's compressed, not stretched?

not true. vertically stretched means horizontally compressed

so y=3x^2 is horizontally compressed, but what you want is horizontally stretched which means vertically compressed. y=(1/3)x^2 is horizontally stretched, yet doesn't include (2, 64) without + 61 1/3

y=(1/3)x^2 +183/3 is horizontally stretched/vertically compressed

or 3y = x^2 +183 to avoid fractions

or x = +/-sqr(3y-183)

to the nearest 10th?

y= .3x^2+61.3

check the answer

64 = .3(2^2) + 61.3

= .3(8+ 61.3 = 2.4+61.3 = 63.7 which is close to 64, so go with it

it may help to draw a rough sketch of the equations

or do an online graph or use a graphing calculator