Roots Intercepts and Turning Points of Quadratic Functions

A quadratic function of a variable $x$ is defined by the expression $f(x) = ax^2 + bx + c$, where $a, b,$ and $c$ represent constants. When plotted on a coordinate plane, the resulting graph is a curve known as a parabola. The orientation of this parabola is determined entirely by the coefficient $a$. If $a > 0$, the parabola opens upwards (a U-shape). If $a < 0$, it opens downwards (an upside-down U-shape).

Roots of quadratic functions graphically

When you plot the graph of $f(x) = ax^2 + bx + c$, the roots of the function are found at the points where the curve intercepts the x-axis. These roots represent the solutions to the quadratic equation $ax^2 + bx + c = 0$. Depending on the specific function, the graph might cross the x-axis twice, touch it once, or not cross it at all.

Intercepts of quadratic functions

The y-axis intercept of the graph $y = ax^2 + bx + c$ occurs when the value of $x$ is zero. Substituting $x = 0$ into the equation gives $y = a(0)^2 + b(0) + c$, which simplifies to $y = c$. On a graph, this is the specific point where the curve crosses the vertical axis.

The x-axis intercepts occur when the value of $y$ is zero, meaning we must solve $ax^2 + bx + c = 0$. There can be 0, 1, or 2 intercepts with the x-axis, which can be identified by looking at the points where the curve meets the horizontal axis.

The turning point of quadratic functions

The turning point is the extreme point of the parabola. It is the minimum (lowest point) if the parabola is U-shaped ($a > 0$) and the maximum (highest point) if the parabola is an upside-down U-shape ($a < 0$). Every quadratic graph is perfectly symmetric, and the vertical line of symmetry always passes directly through the turning point.

Finding the roots of a quadratic function algebraically

To find the roots of $f(x) = ax^2 + bx + c$ algebraically, you must solve the equation $f(x) = 0$. This involves finding the values of $x$ that satisfy $ax^2 + bx + c = 0$, typically through factorisation, the quadratic formula, or completing the square.

Completing the square

Completing the square is a method used to rewrite a quadratic expression. Since $(x + \frac{k}{2})^2 = x^2 + kx + (\frac{k}{2})^2$, we can rearrange this to show that:

$x^2 + kx = (x + \frac{k}{2})^2 - (\frac{k}{2})^2$

where $k$ is a constant. This process expresses the quadratic as a difference of two squares.

Finding the turning point by completing the square

The turning point corresponds to the minimum value of the function if $a > 0$ or the maximum value if $a < 0$. By completing the square, we can determine this point precisely. For the general form $y = ax^2 + bx + c$, we can factor out $a$ and complete the square:

$\frac{y}{a} = x^2 + \frac{b}{a}x + \frac{c}{a} = (x + \frac{b}{2a})^2 - \frac{b^2}{4a^2} + \frac{c}{a}$

To find the turning point, we look for the value of $x$ that makes the squared term equal to zero. Because $(x + \frac{b}{2a})^2$ is always greater than or equal to zero, the turning point occurs when $x + \frac{b}{2a} = 0$, leading to $x = -\frac{b}{2a}$. The corresponding y-coordinate is then $y = -\frac{b^2}{4a^2} + \frac{c}{a}$.

Worked Example: Roots from a graph

Consider the graph of $y = 2x^2 + 6x + 3$ shown below. We want to find the roots correct to 1 decimal place.

On this scale, 5 small squares represent 1 unit, meaning each small square is $0.2$ units. By observing where the curve crosses the x-axis, we find the roots are at approximately $-0.6$ and $-2.4$.

Worked Example: Intercepts

For the function $y = 3x^2 + 2x - 5$, what is the intercept with the y-axis?

The curve intercepts the y-axis when $x = 0$. Substituting this gives: $y = 3(0)^2 + 2(0) - 5 = -5$. Thus, the y-intercept is $-5$.

Worked Example: Turning point from a graph

Look at the graph of $y = 2x^2 + 6x + 3$ again to find its turning point.

The turning point is the location where the curve is parallel to the x-axis. Reading from the graph, the coordinates are $(-1.5, -1.5)$.

Worked Example: Finding roots algebraically

Find the roots of $f(x) = 2x^2 + 5x + 3$.

Set $f(x) = 0$: $2x^2 + 5x + 3 = 0$

Factorising the quadratic gives: $(2x + 3)(x + 1) = 0$

Setting each factor to zero: $2x + 3 = 0$ or $x + 1 = 0$ $x = -\frac{3}{2}$ or $x = -1$

Worked Example: Completing the square practice

1. Complete the square for $x^2 + 4x$: First, halve the coefficient of $x$ to get 2. Since $(x + 2)^2 = x^2 + 4x + 4$, we have: $x^2 + 4x = (x + 2)^2 - 2^2$

2. Complete the square for $x^2 - 6x$: Halve $-6$ to get $-3$. Thus: $x^2 - 6x = (x - 3)^2 - (-3)^2 = (x - 3)^2 - 3^2$

3. Complete the square for $2x^2 + 6x$: First, factor out the 2 so the $x^2$ coefficient is 1: $2(x^2 + 3x)$ Now complete the square for the expression inside the bracket: $2[(x + \frac{3}{2})^2 - (\frac{3}{2})^2] = 2(x + \frac{3}{2})^2 - 2 \times \frac{9}{4} = 2(x + \frac{3}{2})^2 - \frac{9}{2}$

Worked Example: Turning points via completing the square

1. Find the turning point for $y = x^2 + 6x + 7$: Complete the square: $y = (x + 3)^2 - 9 + 7 = (x + 3)^2 - 2$ The turning point occurs when the squared bracket is zero: $x + 3 = 0$, so $x = -3$. Substituting this back gives $y = -2$. The turning point is $(-3, -2)$.

2. Find the turning point for $y = 2x^2 + 6x + 5$: Complete the square: $y = 2(x^2 + 3x) + 5 = 2[(x + \frac{3}{2})^2 - (\frac{3}{2})^2] + 5$ $y = 2(x + \frac{3}{2})^2 - 2 \times \frac{9}{4} + 5 = 2(x + \frac{3}{2})^2 - \frac{9}{2} + 5 = 2(x + \frac{3}{2})^2 + \frac{1}{2}$ This function reaches its minimum when $x + \frac{3}{2} = 0$, which means $x = -\frac{3}{2}$ and $y = \frac{1}{2}$.