Question

On a coordinate plane, a parabola opens up. It goes through (negative 2, 4), has a vertex at (0.25, negative 6), and goes through (2, 0).
Which statements about the graph of the function f(x) = 2x2 – x – 6 are true? Select two options.

The domain of the function is the set of all values x such that x is greater than or equal to one-quarter.
The range of the function is all real numbers.
The vertex of the function is (one-quarter, negative 6 and one-eighth).
The function has two x-intercepts.
The function is increasing over the interval (negative 6 and one-eighth, ∞).

Answer 1

The statements true about the the function f(x) = 2x2 – x – 6 are-

• The vertex of the function is (one-quarter, negative 6 and one-eighth).
• The function has two x-intercepts.

What is vertex of parabola?

The vertex of parabola is the point at the intersection of parabola and its line of symmetry.

Now the given function is,

f(x) = 2x^2 – x – 6

Also, it is given that the vertex is located at (0.25, -6)  and the parabola opens up, the function has two x-intercepts.

Comparing the given function with standard form,

f(x) = a x^2 bx + c

By comprison we get,

a = 2

b = -1

c = -6

Now, x-coordinate of vertex is given as,

x = -b/2a

put the values we get,

x = -(-1)/2*2

or, x = 1/4

Put the value of x in given function, so y-coordinate of the vertex is given as,

f(1/4) = 2(1/4)² - 1/4 - 6

        = -49/6

       = -6 1/8

Hence, The statements true about the the function f(x) = 2x2 – x – 6 are-

• The vertex of the function is (one-quarter, negative 6 and one-eighth).
• The function has two x-intercepts.

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