Question

The graph of the function y=x^2-x-2 is shown below

Answer 1

Answer:

Option D is correct

Step-by-step explanation:

A quadratic equation is in the form of $y =ax^2+bx+c$   ....[1]

then;

the axis of symmetry is given by:

$x = \frac{-b}{2a}$

Vertex = $(\frac{-b}{2a}, f(\frac{-b}{2a})$

As per the statement:

Given the function:

$y=f(x) =x^2-x-2$        ....[2]

On comparing with [1] we have;

a = 1, b = -1 and c = -2

Then;

$x =\frac{-(-1)}{2(1)} = \frac{1}{2}$

Substitute the value of x in f(x) we have;

$f(\frac{1}{2}) = (\frac{1}{2})^2-\frac{1}{2}-2 = \frac{1}{4}-\frac{5}{2} = \frac{1-10}{4} = \frac{-9}{4}$

⇒$f(\frac{1}{2}) = -2\frac{1}{4}$

Vertex =$(\frac{1}{2}, -2\frac{1}{4})$

x-intercept states that the graph crosses the x-axis.

From the graph, the function cuts the x-axis at:

x = -1 and x = 2

x-intercepts = (-1, 0) and (2, 0)

y-intercept states that the graph crosses the y-axis

From the given graph we have;

y = -2

y-intercept = (0, -2)

Therefore,

Vertex =$(\frac{1}{2}, -2\frac{1}{4})$

Axis of symmetry: $x=\frac{1}{2}$

x-intercepts = (-1, 0) and (2, 0)

y-intercept = (0, -2)

Answer 2

The answer to this quistion is D