For the f(x)=ax2+bx+c, the y-coordinate of the is the maximum value of the function when a<0.
Let's identify the values of
a, b, and
c in the given quadratic function.
f(x)=-2x2+4x−1⇕f(x)=-2x2+4x+(-1)
We can see above that
a=-2, b=4, and
c=-1. We will now use these values to find the desired information.
Maximum Value
Since
a=-2 is less than
0, the parabola will open downwards. This means it will have a maximum value, which is given by
f(-2ab). Before we find the value of the function at this point, we need to substitute
a=-2 and
b=4 in
-2ab.
-2ab
▼
Substitute values and evaluate
1
Now we have to calculate
f(1). To do so, we will substitute
1 for
x in the given function.
f(x)=-2x2+4x−1
f(1)=-2(1)2+4(1)−1
f(1)=-2(1)+4(1)−1
f(1)=-2+4−1
f(1)=1
This tells us that the maximum value of the function is
1.
Domain and Range
Unless there are any specified restrictions on the
x-values, the of a quadratic function is all real numbers. Therefore, the domain of this function is all real numbers. Furthermore, since
a=-2 is
less than
0, the is all values less than or equal to the maximum value,
1.
Domain:Range: All real numbers y≤1
Decreasing and Increasing Intervals
Since
a=-2 is
less than
0, the function increases to the left of the maximum value and decreases to the right of the maximum value, which we know occurs at
1.
Increasing Interval:Decreasing Interval: To the left of 1 To the right of 1