Answer :
Points at (1, 3) and (3, 3) and (2, 1)
Explanation:
The original graph (dashed) is an absolute value function meaning f(x) = |x|
<u>Given:</u> y = 2f(x - 2) + 1
f(x - 2) means you plug in x - 2 for x in f(x) function
the new graph equation is: y = 2|x - 2| + 1
- The graph will shift up 1 because +1
- The graph will shift right because -2
- The graph has a scale factor of 2
- The graph is V shaped because it is an absolute value graph
Learn more about Absolute Value here: brainly.com/question/729268
<u>Visual:</u>
Please, use " ^ " to denote exponentiation: p(t) = t^2 + 5t + 6.
To find the critical points, differentiate p(t) with respect to t, set the result = to 0, and then solve the resulting equation for t:
p '(t) = 2t + 5 = 0
Solving for t: 2t = -5, and so t = -5/2. (-5/2, p(-5/2)) is the critical point. That evaluates to (-5/2, -0.25). This happens to be the vertex of a parabola that opens up.
The angles opposite equal sides are also equal. The sum of angles in a triangle is 180. So 2x + 27 + 2x + 27 + 3x + 28 = 180.
7x + 82 = 180
7x = 98
x = 14
Now plug x into each angle to find their measures.
This problem already gave us the slope and y-intercept, so all that's left to do is plug those two values into slope-intercept form.
Slope-Intercept Form: y = mx + b
---m is the slope
---b is the y-intercept
y = 8/3x + 5
Hope this helps!! :)
