SOLUTION: For the quadratic function : f(x) = 2x^2 - 4x -6, find:
a. The x intercepts and the y intercepts
b. THe domain and range
c. The axis of symmetry and the vertex.
d. Find the m
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Question 181400: For the quadratic function : f(x) = 2x^2 - 4x -6, find:
a. The x intercepts and the y intercepts
b. THe domain and range
c. The axis of symmetry and the vertex.
d. Find the minimum value of the function.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
 = 2x^2 - 4x -6)
a. To find the y-intercept, i.e. the value of the function when x = 0, evaluate  = 2(0)^2 - 4(0) -6)
.
To find the x-intercepts, i.e. the values of x that make the value of the function zero, set the function equal to zero and solve the resulting quadratic equation.
b. The domain of a function is that set of numbers for which the function is defined. For this polynomial function, there are no real numbers for which this function is undefined. Therefore the domain is all reals.
Domain: 
The range of a function is that set of numbers that can be the value of the function. Given that this function is a quadratic polynomial with a positive lead coefficient, this is a parabola that opens upward. This means that the vertex of the parabola is a minimum value. The value of the function at the vertex is the minimum value of the range while the maximum value of the range increases without bound.
The x-coordinate of the vertex of a parabola expressed in  = ax^2 + bx + c)
form is given by 
. In this case:
}{2(2)} = 1)
The y-coordinate of the vertex is the value of the function at the The x-coordinate of the vertex, i.e. )
. In this case:
 = 2(1)^2 - 4(1) -6 = -8)
.
Therefore:
Range: 
The axis of symmetry is the vertical line through the vertex. 
We've already calculated the coordinates of the vertex when calculating the lower limit of the range.
The lower limit of the range is the minimum value.
John
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