Substitute “x” with a number
(x1,y1)=(0,6)
(x2,y2)=(6,0)
slope=(0-6)/(6-0)=0
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
Read more about functions at:
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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)
The complete question in the attached figure
we know that
the range is the y-value of the function
so
the lowest y-value on this graph is y = -2.
the answer is
y=-2
The answer is B, 8. 2 exponent 3 is 8 because to find an exponent multiply the base by itself with the exponent number: 2*2*2 which is 8.
