The following diagram shows part of graph of f, where $f(x) = x^2 - x - 2$.
Find both x-intercepts.
For this, $f(x)$ means $y$, should I set $y$ to 0 and solve for the $x$ values?
$0 = x^2 - x - 2$
$2 = x^2 - x$
I got stuck here, and now I’m rethinking that maybe I should do the quadratic formula?
Find the x-coordinate of the vertex.
For this, isn’t there a formula to discover the vertex like with h and k?
3 Answers
$\begingroup$Recall that for $ax^2+bx+c=0$ by quadratic formula the solutions are
$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
and the vertex for the parabola $y=ax^2+bx+c$ is located at
$$x=-\frac b{2a}$$
$\endgroup$ $\begingroup$Find $x$-intercepts of the graph of the function $f(x) = x^2 - x - 2$.
You are correct that the $x$-intercepts are found by setting $y = f(x) = 0$.
Method 1: Factoring.
\begin{align*} f(x) & = 0\\ x^2 - x - 2 & = 0\\ x^2 - 2x + x - 2 & = 0\\ x(x - 2) + 1(x - 2) & = 0\\\ (x + 1)(x - 2) & = 0 \end{align*} By the Zero Product Rule, if $ab = 0$, then $a = 0$ or $b = 0$. Hence, $x + 1 = 0$ or $x - 2 = 0$. \begin{align*} x + 1 & = 0 & x - 2 & = 0\\ x & = -1 & x = 2 \end{align*} Thus, the $x$-intercepts are the point $(-1, 0)$ and $(2, 0)$.
Method 2: Completing the square. \begin{align*} f(x) & = 0\\ x^2 - x - 2 & = 0\\ x^2 - x & = 2\\ x^2 - x + \left(-\frac{1}{2}\right)^2 & = 2 + \left(-\frac{1}{2}\right)^2 \tag{1}\\ x^2 - x + \frac{1}{4} & = 2 + \frac{1}{4}\\ \left(x - \frac{1}{2}\right)^2 & = \frac{9}{4} \tag{2}\\ \left|x - \frac{1}{2}\right| & = \frac{3}{2} \tag{3}\\ x - \frac{1}{2} & = \pm \frac{3}{2} x & = \frac{1}{2} \pm \frac{3}{2} \end{align*} \begin{align*} x & = \frac{1}{2} + \frac{3}{2} & x & = \frac{1}{2} - \frac{3}{2}\\ x & = 2 & x & = -1 \end{align*} Again, we obtain the $x$-intercepts $(-1, 0)$ and $(2, 0)$.
(1) Since the leading coefficient is $1$, take half the coefficient of the linear term, which in this case is $-1$, square it, and add the result to both sides of the equation. This produces a perfect square on the left-hand side since $x^2 + 2bx + b^2 = (x + b)^2$. Here $2b = -1 \implies b = -1/2$.
(2) Factor the perfect square on the left-hand side of the equation.
(3) Take square roots. Notice that if the number on the right-hand side is negative in the preceding step is negative, there is no real-valued solution, which means there would be no $x$-intercepts.
Method 3: Apply the Quadratic Formula.
To apply the Quadratic Formula, the equation must be written in the form $ax^2 + bx + c = 0$. \begin{align*} f(x) & = 0 x^2 - x - 2 & = 0 \end{align*} Here, $a = 1$, $b = -1$, and $c = -2$. Hence, \begin{align*} x & = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\ & = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)}\\ & = \frac{1 \pm \sqrt{1 + 8}}{2}\\ & = \frac{1 \pm \sqrt{9}}{2}\\ & = \frac{1 \pm 3}{2} \end{align*} \begin{align*} x & = \frac{1 + 3}{2} & x & = \frac{1 - 3}{2}\\ & = \frac{4}{2} & x & = \frac{-2}{2}\\ & = 2 & x & = -1 \end{align*} Again, we obtain the $x$-intercepts $(-1, 0)$ and $(2, 0)$.
Note: Completing the square and applying the Quadratic Formula always work. Factoring may not. However, in this case, factoring is simpler.
Find the vertex of the graph of the function $f(x) = x^2 - x - 2$.
If we express the function in the form $f(x) = a(x - h)^2 + k$, the vertex is $(h, k)$.
Method 1: We complete the square. \begin{align*} f(x) & = x^2 - x - 2\\ & = (x^2 - x) - 2\\ & = \left[x^2 - x + \left(-\frac{1}{2}\right)^2\right] - 2 - \left(-\frac{1}{2}\right)^2 \tag{1}\\ & = \left(x^2 - x + \frac{1}{4}\right) - 2 - \frac{1}{4} \tag{2}\\ & = \left(x - \frac{1}{2}\right)^2 - \frac{9}{4} \end{align*} Hence, the vertex is $\left(\dfrac{1}{2}, -\dfrac{9}{4}\right)$.
(1) Since the leading coefficient is $1$, take half the coefficient of the linear term, which in this case is $-1$, square it, add the result inside the parentheses and subtract it outside the parentheses.
(2) Factor the perfect square.
Method 2: We use the formula $$x = -\frac{b}{2a}$$ to find the axis of symmetry, then substitute this value for $x$ to find the $y$-coordinate of the vertex.
$f(x) = x^2 - x - 2$, so $a = 1$, $b = -1$, and $c = -2$. Thus, the axis of symmetry is $$x = -\frac{b}{2a} = -\frac{-1}{2 \cdot 1} = \frac{1}{2}$$ Substituting $1/2$ for $x$ in the function gives $$f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} - 2 = \frac{1}{4} - \frac{1}{2} - 2 = -\frac{9}{4}$$ which gives the vertex $\left(\dfrac{1}{2}, \dfrac{9}{4}\right)$.
The formula for the axis of symmetry is obtained by completing the square on $f(x) = ax^2 + bx + c$.
\begin{align*}
f(x) & = ax^2 + bx + c\\ & = a\left(x^2 + \frac{b}{a}\right) + c\\ & = a\left[x^2 + \frac{b}{a}x + \left(\frac{1}{2} \cdot \frac{b}{a}\right)^2\right] - a\left(\frac{1}{2} \cdot \frac{b}{2a}\right)^2 + c\\ & = a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}\right) - \frac{b^2}{4a} + c\\ & = a\left(x + \frac{b}{a}\right)^2 + \frac{-b^2 + 4ac}{4a}
\end{align*}
which gives axis of symmetry
$$x = -\frac{b}{2a}$$
and vertex
$$\left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right)$$
$x^2-x-2=(x-2)(x+1)=0$
So $(x=2, y=0)$ and $(x=-1, y=0)$ are x-intercepts.
For vertex as you see the minimum of the function is the vertex so we have:
$y'=2x-1=0$ ⇒$x=1/2 $
$y=(1/2)^2-1/2-2=-9/4$
So vertex is: $(x=1/2, y=-9/4)$
$\endgroup$
