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3 years ago

6

Jim needs 3 1/2 of burlap to cover his bulletin board. His brother gives him 1 1/6 yards. How much more does Jim need to purchas

e?

1 answer:

ANTONII [103]3 years ago

3 0

1 2/3 yards
Hope this helps!
Vote my brainliest!

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If the parent function f(x) = x^2 is fatter translated 11 units to the left, then translated 5 units down, write the resulting f

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The quadratic function given by:

$f(x)=a(x-h)^2+k, \ \ \ a\neq 0$

is in vertex form. The graph of $f$ is a parabola whose axis is the vertical line $x=h$ and whose vertex is the point $(h, k)$ . So:

To translate the graph of a function to the right, left, upward or downward we have:

$For \ a \ positive \ real \ number \ c. \ \mathbf{Vertical \ and \ horizontal \ shifts} \\ in \ the \ graph \ of \ y=f(x) \ are \ represented \ as \ follows:\\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{upward}: \\ g(x)=f(x)+c \\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{downward}: \\ g(x)=f(x)-c \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ \mathbf{right}: \\ g(x)=f(x-c) \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ \mathbf{left}: \\ g(x)=f(x+c)$

By knowing this things, we can solve our problem as follows:

FIRST.

Translating <em>11 units to the left:</em>

$g(x)=f(x+11) \\ \\ \therefore g(x)=(x+11)^2$

Then translating<em> 5 units down:</em>

$g(x)=f(x)-c \\ \\ \therefore g(x)=(x+11)^2-5$

Since the new function is fatter, the factor we need to multiply the term $(x+11)^2$ <em>must be</em> less than 1, to make the graph fatter. So, according to our options, there are two factors 1/2 and 2.

<em>Therefore, the right answer is </em><em>b. f(x) = 1/2(x + 11)^2 - 5</em>

SECOND.

Translating <em>8 units to the right:</em>

$g(x)=f(x-8) \\ \\ \therefore g(x)=(x-8)^2$

Then translating<em> 1 unit down:</em>

$g(x)=f(x)-c \\ \\ \therefore g(x)=(x-8)^2-1$

As explained in the previous case, there are two factors 1/3 and 3, so we choose the first one.

<em>Therefore, the right answer is </em><em>a. g(x) = 1/3(x - 8)^2 - 1</em>

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