Question 984492: What is the smallest value, to 2 decimal places, in the image set of the function
f(x) = ax^2 + bx + c
when a = 2, b = 4 and c = 121? Answer by Edwin McCravy(20060) (Show Source):
The smallest value of a quadratic function with a positive leading coefficient
occurs at the vertex, of which the x-coordinate is . The minimum
value is the y-coordinate of the function when that x-coordinate is substituted
for x in the function.
The largest value of a quadratic function with a negative leading coefficient
occurs at the vertex, of which the x-coordinate is . The maximum
value is the y-coordinate of the function when that x-coordinate is substituted
for x in the function.
Instead of doing your problem for you, I'll do one exactly in every step and
detail like yours. You can use it as a model to do yours.
f(x) = ax^2 + bx + c
when a = 3, b = 12 and c = 127
f(x) = 3x^2 + 12x + 127
The x-coordinate of the vertex is
= = =
We find the y-ccordinate by substituting -2 for x in
f(x) = 3x^2 + 12x + 127
f(2) = 3(-2)^2 + 12(-2) + 127
f(2) = 3(4) - 24 + 127
f(2) = 12 - 24 + 127
f(2) = 115
So the minimum value is 115.
Now do yours exactly the same way.
Edwin
