Quadratics Revisited

Previously, we have seen that given $a > 0$ , the graph of $y = ax^2$ looks like the diagram below.

For any point $P(x, y)$ on the graph, it has the same distance to the (focal) point $F(0, 1/(4a))$ and the (directrix) line $y = -1/(4a)$ . Hence, its shape is known as a parabola. In particular, it has a minimum point $(0, 0)$ .

If we shifted the graph rightwards by $h$ and upwards by $k$ , we recover a more general parabola with minimum point $(h, k)$ .

Of course, if $a < 0$ , then the whole graph gets “flipped” to the downside:

The diagram above shows the graph of $y = -x^2 + 4$ .

Example 1. Determine the $y$ -intercept of the quadratic graph $y = -x^2 + 4$ .

Solution. To determine the $y$ -intercept of the quadratic graph, we need to find the intersection of the graph $y = -x^2 + 4$ and the $y$ -axis, whose equation is $x = 0$ . Since the $x$ -values should match, we can substitute the latter equation into the former:

$y = -0^2 + 4 = 4.$

Therefore, our $y$ -intercept is $(0, 4)$ . No surprises there.

Example 2. Determine the two $x$ -intercepts of the quadratic graph $y = -x^2 + 4$ .

Solution. To determine the two $x$ -intercepts of the quadratic graph, we need to find the intersections of the graph $y = -x^2 + 4$ and the $x$ -axis, whose equation is $y = 0$ . Since the $y$ -values should match, we can substitute the latter equation into the former:

$0 = -x^2 + 4 \quad \Rightarrow \quad x^2 = 4.$

It is tempting at this stage to write $x = \sqrt 4 = 2$ so that our $x$ -intercept is $(2, 0)$ . However, this solution is only partially correct. Just like any solution in Singapore or US politics.

Notice in the graph that we have a second $x$ -intercept to the left of the $y$ -axis. This is because $x$ could have been a negative number as well. Notice that

$2^2 = 4\quad \text{and}\quad (-2)^2 = 4$

are both correct equations. (Furthermore, these are the only two correct equations). Therefore, if $x^2 = 4$ , we must conclude either possibility, namely, that $x = -2$ or $x = 2$ . To abbreviate, we use the $\pm$ notation: $x = \pm 2$ . This notation states that either $-2$ or $+2$ are plausible values that $x$ represents.

To answer our original question, since $x = \pm 2$ and $y = 0$ in both cases, there are two $x$ -intercepts: $(-2, 0)$ and $(2, 0)$ .

Example 3. Determine the $x$ -intercept(s) of the quadratic graph $y = x^2 - 4$ .

Solution. To determine the two $x$ -intercepts of the quadratic graph, we need to find the intersections of the graph $y = x^2 - 4$ and the $x$ -axis, whose equation is $y = 0$ . Since the $y$ -values should match, we can substitute the latter equation into the former:

$x^2 - 4 = 0 \quad \iff \quad 0 = -x^2 + 4.$

By Example 2, we have the solutions $x = \pm 2$ . Therefore, there are two $x$ -intercepts: $(-2, 0)$ and $(2, 0)$ .

Remark 1. Different graphs can yield the same roots.

Example 4. Determine the $x$ -intercept(s) of the quadratic graph $y = x^2 - 2x + c$ , where $c = -3, -1, 1, 2$ respectively.

Solution. We recall that $(x-1)^2 = x^2 - 2x + 1$ . Therefore,

$\begin{aligned} y = x^2 - 2x + c = (x-1)^2 - 1 + c. \end{aligned}$

To find the $x$ -intercepts of the graph, we set $y = 0$ :

$\begin{aligned} (x-1)^2 - 1 + c &= 0 \\ (x-1)^2 &= 1 - c. \end{aligned}$

Now we analyse this equation case-by-case.

If $c = -3$ , then $(x-1)^2 = 1-(-3) = 4$ . Taking square roots,

$x-1 = \pm \sqrt 4 = \pm 2 \quad \Rightarrow \quad x = 1 \pm 2.$

Hence, $x = -1$ or $x = 3$ , yielding the $x$ -intercepts $(-1, 0)$ and $(3, 0)$ .

If $c = -1$ , then $(x-1)^2 = 1-(-1) = 2$ . Taking square roots,

$x-1 = \pm \sqrt 2 \quad \Rightarrow \quad x = 1 \pm \sqrt 2.$

Hence, $x = 1 - \sqrt 2$ or $x = 1 + \sqrt 2$ , yielding the $x$ -intercepts $(1 - \sqrt 2, 0)$ and $(1 + \sqrt 2, 0)$ .

If $c = 1$ , then $(x-1)^2 = 1-1 = 0$ . Taking square roots,

$x-1 = 0 \quad \Rightarrow \quad x = 1.$

Hence, $x = 1$ , yielding only one $x$ -intercept $(1, 0)$ .

If $c = 3$ , then $(x-1)^2 = 1-3 = -2$ . Now, if $x$ is a real number, then $(x-1)$ is a real number too, which implies that $(x-1)^2 \geq 0 > -1$ . Therefore, $(x-1)^2 \neq -2$ . This can only mean that the equation $(x-1)^2 = -2$ has no real solutions, and therefore, the graph of $y = x^2 - 2x + 2$ no $x$ -intercepts.

The diagram below shows the graphs of all four cases.

Example 5. Given real numbers $h$ and $k$ , determine the number of $x$ -intercepts of the quadratic graph $y = (x-h)^2 + k$ , and determine them in terms of $h$ and $k$ if they exist.

Solution. Repeating the calculations in Example 4, we set $y = 0$ :

$(x-h)^2 + k = 0 \quad \Rightarrow \quad (x-h)^2 = -k.$

There are three cases: $k < 0$ , $k = 0$ , or $k > 0$ .

In the case $k < 0$ , $-k > 0$ , so that $\sqrt{-k} > 0$ . Hence,

$(x-h)^2 = -k \quad \Rightarrow \quad x = h \pm \sqrt{-k}.$

Therefore, the graph has two $x$ -intercepts $(h - \sqrt{-k}, 0)$ and $(h + \sqrt{-k}, 0)$ .

In the case $k = 0$ ,

$(x-h)^2 = -k \quad \Rightarrow \quad x = h \pm \sqrt{-k} = h \pm \sqrt{-0} = h.$

Therefore, the graph has only one $x$ -intercept $(h, 0)$ .

In the case $k > 0$ , $-k < 0$ , so that

$(x-h)^2 \geq 0 > -k.$

Therefore, the equation $(x-h)^2 = -k$ has no real solutions, and the graph has no $x$ -intercepts.

Remark 2. The line $x = h$ is called the line of symmetry of the graph, and the point $(h, k)$ is the turning point of the graph.

Definition 1. We say that $x_0$ is a real number solution of the quadratic equation $ax^2 + bx + c = 0$ if $ax_0^2 + bx_0 + c = 0$ . Equivalently,

$x_0$ is a real number solution of the equation $ax^2 + bx + c = 0$ ,
$x_0$ is a root of the function $ax^2 + bx + c$ ,
$(x_0, 0)$ is an $x$ -intercept of the graph $y = ax^2 + bx + c$ .

Theorem 1. Let $ax^2 + bx + c = 0$ be a quadratic equation, where $a \neq 0$ . Define the discriminant $\Delta$ of the quadratic equation by

$\Delta := b^2 - 4ac.$

Then the equation $ax^2 + bx + c = 0$ has

$2$ real number solutions if $\Delta > 0$ ,
$1$ real number solution if $\Delta = 0$ , and
$0$ real number solutions if $\Delta < 0$ .

In the first two cases, the roots $x_{\pm}$ are given by the quadratic formula:

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \equiv \frac{-b \pm \sqrt{\Delta}}{2a}.$

Proof Sketch. We first leave it as an exercise to find constants $h, k$ in terms of $a,b, c$ such that

$\displaystyle y = ax^2 + bx + c \quad \iff \quad \frac ya = (x-h)^2 + k.$

This process is called completing the square, and done correctly, should yield the results

$\displaystyle h = - \frac b{2a},\quad k = -\frac{\Delta}{(2a)^2}.$

Then set $y = 0$ and apply the calculation in Example 5. In the case $a > 0$ and $\Delta \geq 0$ ,

$\displaystyle x = -h \pm \sqrt{-k} = -\frac{b}{2a} \pm \sqrt{\frac{\Delta}{(2a)^2}} = \frac{ -b \pm \sqrt{ \Delta } }{2a}.$

Once again, the square roots make their return. In Example 4, the equation $x^2 - 2x - 1 = 0$ has the roots given by $x = 1 \pm \sqrt 2$ . We will explore expressions of this form in the next post.

—Joel Kindiak, 24 Oct 25, 1256H

KindiakMath

Quadratics Revisited

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Quadratics Revisited

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