For a function to be continuous at an x-value, say -17, you need to make sure two things line up:
The limit from the left equals the limit from the right.

This limit equals the functions value.

The left hand limit involves the first piece, f(x) = 20x + 1:
![\begin{aligned} \lim_{x \to -17^{-}} f(x) &= \lim_{x \to -17^{-}} (20x+1)\\[0.5em]&= 20(-17)+1\\[0.5em]&= -339\endaligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20%5Clim_%7Bx%20%5Cto%20-17%5E%7B-%7D%7D%20f%28x%29%20%26%3D%20%20%5Clim_%7Bx%20%5Cto%20-17%5E%7B-%7D%7D%20%2820x%2B1%29%5C%5C%5B0.5em%5D%26%3D%20%20%2020%28-17%29%2B1%5C%5C%5B0.5em%5D%26%3D%20%20%20-339%5Cendaligned%7D)
The right hand limit invovles the second piece, f(x) = -10x^2:
![\begin{aligned} \lim_{x \to -17^{+}} f(x) &= \lim_{x \to -17^{+}} (-10x^2)\\[0.5em]&= -10\cdot (-17)^2\\[0.5em]&= -2890\endaligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20%5Clim_%7Bx%20%5Cto%20-17%5E%7B%2B%7D%7D%20f%28x%29%20%26%3D%20%20%5Clim_%7Bx%20%5Cto%20-17%5E%7B%2B%7D%7D%20%28-10x%5E2%29%5C%5C%5B0.5em%5D%26%3D%20%20%20-10%5Ccdot%20%28-17%29%5E2%5C%5C%5B0.5em%5D%26%3D%20%20%20-2890%5Cendaligned%7D)
Since the two one-sided limits don't match, the function is not continuous at x=-17.
9514 1404 393
Answer:
Step-by-step explanation:
Angles A and P are marked congruent; angles B and Q are marked congruent, so the triangles are similar by the AA similarity postulate. The similarity statement can be written ...
ΔABC ~ ΔPQR by AA similarity
__
The ratio of sides of PQR to ABC is the ratio QR/BC = 12/6 = 2. That is, each side in the larger triangle is 2 times the length of the corresponding smaller side.
PQ = 2·AB = 2·4 = 8
PR = 2·AC = 2·7 = 14
The side lengths of interest are ...
PQ = 8, PR = 14
In this we know all three zeros and one point from which the graph pass.
So we will let specific cubic polynomial function of the form

As we know zeros are that point where we will get value of function equal to zero. So it is basically in form

SO in given question zeros are (2 , 0) , (3, 0) and (5,0)
So we can say

So required equation is

Now we have one point (0 , -5) from which graph passes.
So we say at x = 0 , f(x) = -5



So required equation of cubic polynomial is

For finding y - intercept we simply plugin x = 0 in given equation.
As we know at x = 0 , value of function is -5.
So y - intercept is -5.
Find a common denominator that 8 and 3 can go into.. which is 24
5/8 = 1/3
15/24 - 8/24
7/24
Hope this helped