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3 years ago

Describe the continuity of the graphed function

Mathematics

1 answer:

marusya05 [52]3 years ago

5 0

we know that

Any function f(x) is continuous at x=a only if

$\lim_{x \to a-} f(x) = \lim_{x \to a+} f(x)=f(a)$

We can see that this curve is smooth everywhere except at x=3

so, we will check continuity at x=3

Left limit is:

$\lim_{x \to 3-} f(x) = -\infty$

Right limit is:

$\lim_{x \to 3+} f(x) = +\infty$

Functional value:

$f(3)= DNE$

we can see that left limit is not equal to right limit

so, limit does not not exist

so, this function is discontinuous at x=3

Since, limit does not exist

so, there will be non-removal discontinuity at x=3

so, option-C........Answer

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

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 11 units to the left:

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

Then translating 5 units down:

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

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

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

SECOND.

Translating 8 units to the right:

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

Then translating 1 unit down:

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

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

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

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