hello! i just needed help to verify or falsify my answer and any explanation as to why would be extremely helpful! thanks ahead of time.
y=(x^2)-6x+10
a. at x=3
b. at x=0
c. at x=1
d. at x=-3
my answer was D.
thanks again!
just realized i didn't even specify the directions, haha. it states 'Find the horizontal tangents of the curve.'
Huh?
There are values of y for each of those values of x
at x = -3
y = 9 + 18 + 10 = 37
at x = 3
y = 9 -18 + 10 = 1
what is the question?
Oh
y' = 2x-6
for 0, x = 3
Yeah, i did forget to include that. I put it in a comment above though! Thanks again and sorry about that. 'Find the horizontal tangents of the curve."
You do not really need calculus. It is a parabola, find the vertex
y=(x^2)-6x+10
x^2 - 6 x = y - 10
x^2 - 6 x - 9 = y-19
(x-3)^2 = (y-19)
vertex at (3 , 19)
thank you!!
also had another quick question. when something asks for y'', is there a name for that type of derivative? like what does y'' mean? thanks again. sorry for having so many questions.
Hello! I'd be happy to help you verify your answer for the given equation and explain the process.
The equation you have is y = x^2 - 6x + 10.
To check whether a given value of x will yield the correct value of y, you need to substitute that x-value into the equation and evaluate it.
Let's go through each option:
a. At x = 3:
Substituting x = 3 into the equation:
y = (3^2) - 6(3) + 10 = 9 - 18 + 10 = 1
So, y = 1 when x = 3.
b. At x = 0:
Substituting x = 0 into the equation:
y = (0^2) - 6(0) + 10 = 0 - 0 + 10 = 10
So, y = 10 when x = 0.
c. At x = 1:
Substituting x = 1 into the equation:
y = (1^2) - 6(1) + 10 = 1 - 6 + 10 = 5
So, y = 5 when x = 1.
d. At x = -3:
Substituting x = -3 into the equation:
y = ((-3)^2) - 6(-3) + 10 = 9 + 18 + 10 = 37
So, y = 37 when x = -3.
Therefore, your answer of D, y = 37 at x = -3, is correct! Keep in mind that this process can be used to verify any answer by substituting the given x-values into the equation.