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4 years ago

Consider the following.f(x)=74 – x2Find the x-values at which f is not continuous. Which of the discontinuities are removable?

Mathematics

1 answer:

tensa zangetsu [6.8K]4 years ago

6 0

Answer

given,

f(x) = 74 - x²

A function is continuous when it is differential at all the points.

$f'(x) =\dfrac{d}{dx}(74-x^2)$

$f'(x) =- 2 x$

so, the given function is continuous for all the value of x.

We should know that if a function is quadratic then it is differential at all the point. And if it is differential then it is definitely continuous.

Hence, there is no discontinuous point present in the equation.

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Write the equation in vertex form and then find the vertex, focus, and directrix of a parabola with equation x=y^2+18y-2

kakasveta [241]

Answer:

Part 1) The vertex is the point (-83,-9)

Part 2) The focus is the point (-82.75,-9)

Part 3) The directrix is $x=-83.25$

Step-by-step explanation:

step 1

Find the vertex

we know that

The equation of a horizontal parabola in the standard form is equal to

$(y - k)^{2}=4p(x - h)$

where

p≠ 0.

(h,k) is the vertex

(h + p, k) is the focus

x=h-p is the directrix

In this problem we have

$x=y^{2} +18y-2$

Convert to standard form

$x+2=y^{2} +18y$

$x+2+81=y^{2} +18y+81$

$x+83=(y+9)^{2}$

This is a horizontal parabola open to the right

(h,k) is the point (-83,-9)

The vertex is the point (-83,-9)

step 2

we have

$x+83=(y+9)^{2}$

<em>Find the value of p</em>

$4p=1$

$p=1/4$

<em>Find the focus</em>

(h + p, k) is the focus

substitute

(-83+1/4,-9)

The focus is the point (-82.75,-9)

step 3

Find the directrix

The directrix of a horizontal parabola is

$x=h-p$

substitute

$x=-83-1/4$

$x=-83.25$

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Answer:

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Answer:

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