The value of the function h(x + 1) is -x^2 - x + 1
<h3>How to evaluate the function?</h3>
The equation of the function is given as:
h(t) =-t^2 + t + 1
The function is given as:
h(x + 1)
This means that t = x + 1
So, we substitute t = x + 1 in the equation h(t) =-t^2 + t + 1
h(x + 1) =-(x + 1)^2 + (x + 1) + 1
Evaluate the exponent
h(x + 1) =-(x^2 + 2x + 1) + x + 1 + 1
Expand the brackets
h(x + 1) = -x^2 - 2x - 1 + x + 1 + 1
Evaluate the like terms
h(x + 1) = -x^2 - x + 1
Hence, the value of the function h(x + 1) is -x^2 - x + 1
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<u>Complete question</u>
Consider the following function definition, and calculate the value of the function
h(t) = −t2 + t + 1 h(x + 1)
<span>6( 5-8x) +12= -54
30-48x+12=-54 this comes from distributing(multiplying) 6x5 and 6 times -8x
30+12-48x=-54 you have to add the coefficients 30+12
42-48x=-54
-42 -42
-----------------------
-48x=-96 divide by -48
x=2</span>
Answer:
1/5
Step-by-step explanation:
A reflection through the axis and is given by the following transformation rule:
(x, y) -------> (-x, y)
We have the following point:
C = (5, 3)
Applying the transformation rule we have:
(5, 3) -------> (-5, 3)
Therefore, C' is given by:
C '= (- 5, 3)
Answer:
(-5, 3)
