I need help with this problem.
2 answers:
Answer:
(1,4)
Step-by-step explanation:
Let's this of which is it's parent function.
How do we get to from there?
It has been reflected about the a-axis because of the - in front of the square root.
It has been shifted right 1 unit because of the -(1) in the square root.
It has been moved up 4 units because of the +4 outside the square root.
In general:
has the following transformations from the parent:
Moved right h units if h is positive.
Moved left h units if h is negative.
Moved up k units if k is positive.
Moved down k units if k is negative.
If is positive, it has not been reflected.
If is negative, it has been reflected about the x-axis.
also tells us about the stretching factor.
The parent function has a starting point at (0,0). Where does this point move on the new graph?
It new graphed was the parent function but reflected over x-axis and shifted right 1 unit and moved up 4 units.
So the new starting point is (0+1,0+4)=(1,4).
Answer:
<h2>(1, 4)</h2>Step-by-step explanation:
You must specify the domain of the function.
We know: There is no square root of the negative number.
Therefore
x - 1 ≥ 0 <em>add 1 to both sides</em>
x ≥ 1
The first argument for which the function exists is the number 1.
We will calculate the function value for x = 1.
Put x = 1 to the equation of the function:
Therefore the starting point of the graph of given function is (1, 4).
You might be interested in
The smallest positive integer that the intermediate value theorem guarantees a zero exists between 0 and a is 3.
What is the intermediate value theorem?
Intermediate value theorem is theorem about all possible y-value in between two known y-value.
x-intercepts
-x^2 + x + 2 = 0
x^2 - x - 2 = 0
(x + 1)(x - 2) = 0
x = -1, x = 2
y intercepts
f(0) = -x^2 + x + 2
f(0) = -0^2 + 0 + 2
f(0) = 2
(Graph attached)
From the graph we know the smallest positive integer value that the intermediate value theorem guarantees a zero exists between 0 and a is 3
For proof, the zero exists when x = 2 and f(3) = -4 < 0 and f(0) = 2 > 0.
<em>Your question is not complete, but most probably your full questions was</em>
<em>Given the polynomial f(x)=− x 2 +x+2 , what is the smallest positive integer a such that the Intermediate Value Theorem guarantees a zero exists between 0 and a ?</em>
Thus, the smallest positive integer that the intermediate value theorem guarantees a zero exists between 0 and a is 3.
Learn more about intermediate value theorem here:
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