For this case we have the following equation:
y = x2-4x + 3
Deriving we have the following equation:
y '= 2x-4
We equal zero and clear x:
2x-4 = 0
x = 4/2
x = 2
Substituting in the given equation we have:
y = (2) ^ 2-4 (2) +3
y = 4-8 + 3
y = -1
The vertex will be the ordered pair:
(x, y) = (2, -1)
Answer:
(x, y) = (2, -1)
option B
Answer:
13 and -14 satisfy this condition
Step-by-step explanation:
Let's represent that number as x
and the square of x is x^2
So,
x + x^2 = 182
Subtract 182 from both sides
x + x^2 - 182 = 182 - 182
x + x^2 - 182 = 0
rearrange the quadratic equation
x^2 + x -182 = 0
let's use the quadratic formula
or 
a = 1, b = 1, c = -182
or 
or 
or 
or 
or 
13 or - 14
Lets check
13 + 13^2 = 13 + 169
= 182
Also,
-14 + (-14^2) = -14 + 196
= 182
We can set it up like this, where <em>s </em>is the speed of the canoeist:

To make a common denominator between the fractions, we can multiply the whole equation by s(s-5):
![s(s-5)[\frac{18}{s} + \frac{4}{s-5} = 3] \\ 18(s-5)+4s=3s(s-5) \\ 18s - 90+4s=3 s^{2} -15s](https://tex.z-dn.net/?f=s%28s-5%29%5B%5Cfrac%7B18%7D%7Bs%7D%20%2B%20%5Cfrac%7B4%7D%7Bs-5%7D%20%3D%203%5D%20%5C%5C%2018%28s-5%29%2B4s%3D3s%28s-5%29%20%5C%5C%2018s%20-%2090%2B4s%3D3%20s%5E%7B2%7D%20-15s)
If we rearrange this, we can turn it into a quadratic equation and factor:

Technically, either of these solutions would work when plugged into the original equation, but I would use the second solution because it's a little "neater." We have the speed for the first part of the trip (9 mph); now we just need to subtract 5mph to get the speed for the second part of the trip.

The canoeist's speed on the first part of the trip was 9mph, and their speed on the second part was 4mph.