The vertex of the function y=-3(x - 1)² +2 is?

1 answer:

7 0

Formula: f(x) = a(x – h)2 + k

transformation: 1 unit right 2 units up

vertex: (h,k) -> (1,2) it’s positive because it has to be the opposite

maximum/minimum: maximum

range: (-∞, ∞)
domain: y ≥2

You might be interested in

The slope-intercept form of the equation of a line that passes through point (-3, 8) is y = -2/3x + 6. What is the point-

Elena-2011 [213]

$y = \stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{3}}x+6\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so hmmm then we know the slope of that line is -2/3, so we're really looking for the point-slope form of a line with a slope of -2/3 and that passes through (-3 , 8)

$(\stackrel{x_1}{-3}~,~\stackrel{y_1}{8})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{8}=\stackrel{m}{-\cfrac{2}{3}}(x-\stackrel{x_1}{(-3)})\implies y-8=-\cfrac{2}{3}(x+3)$

3 0

1 year ago

Let $f(x)$ be the real-valued function defined for all real $x$ except for $x = 0$ and $x = 1$ and satisfying the functional equ

Mademuasel [1]

We have to find the values of F.
In this case. F is unlikely to be a polynomial.
But the problem is, we can’t calculate the values of F directly.
There is no real value of x for which x = x−1 x because F isn’t defined at 0 or 1. so,
substituting x = 2.
F(2) + F(1/2) = 3.

Substitute, x = 1/2
F(1/2) + F(−1) = −1/2.
We still are not getting the required value,
therefore,
Substitute x = −1

As, F(2) +F(−1) = 0.
now we have three equations in three unknowns, which we can solve.
It turns out that:
F(2) = 3/4
F(3) = 17/12
F(4) = 47/24
and
F(5) = 99/40

Setting
g(x) = 1 − 1/x
and using
2 → 1/2
to denote
g(2) = 1/2
we see that :
x → 1 - 1/x → 1/(1-x) →x

so that:
g(g(g(x))) = x.

Therefore, whatever x 6= 0, 1 we start with, we will always get three equations in the three “unknowns” F(x), F(g(x)) and F(g(g(x))).
Now solve these equations to get a formula for F(x)

As,
h(x) = (1+x)/(1−x)
which satisfies h(h(h(h(x)))) = x

Now, mapping x to h(x) corresponds to rotating the circle by ninety degrees.