Answer
Graph $B$
Work Step by Step
We need to write $f(x)$ in the form
$y = a \cdot (x-h)^2 + k$
where $(h,k)$ is the vertex of the parabola.
So, we write $y = f(x)$, then:
$y = x^2 - 2 \cdot x - 3$
Taking $-3$ to the other side positive, we get:
$y+3 = x^2 - 2 \cdot x$
Put $1$ in evidence on the right hand side, we have:
$y+3 = 1 \cdot (x^2 - 2 \cdot x)$
The coefficient of $x$ is $-2$, then:
$\frac{-2}{2} = -1$ and $(-1)^2 = 1$, so we have to add $1$ to both sides of the equality, in order to get:
$y+3 + (+ 1)=1 \cdot (x^2-2 \cdot x+1)$
Now, we factorize the right hand side, to get:
$y + 3 + 1 = 1 \cdot (x-1)^2$
$y + 4 = 1 \cdot (x-1)^2$
Then:
$y = 1 \cdot (x-1)^2 - 4$
$y= 1 \cdot (x-1)^2 + (-4)$
So, is in the form that we wanted and:
Vertex $(1,-4)$
Then, $a = 1 > 0$, and the parabola opens upward and the symmetry axis is the line $x=1$.
Now, let's find $x$-intercepts. Take:
$y = 0$
$x^2 -2 \cdot x - 3 = 0$
Factorizing, we look two number such that its sum is $-2$ and its product is $-3$, looking to a pair with opposite signs. We think:
$1$ and $-3$
$-1$ and $3$
The right one is $1$ and $-3$, because:
$1+(-3) = -2$
And
$1 \cdot (-3) = -3$
So
$x^2 + 1 \cdot x - 3 \cdot x - 3 =0$
$=x \cdot (x+1)-3 \cdot (x+1) =0$
$(x-3) \cdot (x+1)=0$
Therefore, the roots are:
$a = 3$
$b=-1$
So, the x-intercepts are in the points
$(3,0)$ and $(-1,0)$
Finally, we take $x=0$ in order to calculate the $y$-intercept. Then:
$f(0) = 0^2 - 2 \cdot 0 - 3 = -3$
Then, the $y$-intercept is in the point $(0,-3)$.
Therefore the correct choice is Graph $B$.
