Roots Intercepts and Turning Points of Quadratic Functions

A quadratic function is a mathematical expression of the form $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constant values. When plotted on a coordinate plane, the resulting graph is a curve called a parabola. The orientation of this parabola is determined by the sign of $a$:

1. If $a > 0$, the parabola is U-shaped (concave up), opening upwards.
2. If $a < 0$, the parabola is an upside down U-shape (concave down), opening downwards.

Graphical Interpretation of Roots and Intercepts

Roots of Quadratic Functions

If you plot the graph of a quadratic function $f(x) = ax^2 + bx + c$, the points where the curve crosses or touches the $x$ axis are the roots of the function. These roots are the solutions to the equation $ax^2 + bx + c = 0$.

Intercepts

The $y$ intercept of a quadratic graph $y = ax^2 + bx + c$ occurs when $x = 0$. Substituting zero into the equation yields $y = c$. This is the point where the curve crosses the vertical axis.

The $x$ intercepts occur when $y = 0$. Depending on the specific function, there may be zero, one, or two intercepts with the $x$ axis. These points correspond directly to the real roots of the quadratic.

The Turning Point

The turning point is the vertex of the parabola. It is the minimum point if $a > 0$ and the maximum point if $a < 0$. Quadratic graphs are always symmetrical, and the line of symmetry is a vertical line that passes directly through this turning point.

Worked Example: Graphical Identification

The following diagram shows a section of the graph of $y = 2x^2 + 6x + 3$.

Finding Roots: To find the roots to 1 decimal place, we look at where the curve crosses the $x$ axis. Given that 5 small squares represent 1 unit, each small square is $0.2$. From the graph, we can see the roots are approximately $-0.6$ and $-2.4$.

Finding Intercepts: For $y = 3x^2 + 2x - 5$, the $y$ intercept occurs when $x = 0$. Thus, $y = 0 + 0 - 5 = -5$.

Finding the Turning Point: Using the same graph of $y = 2x^2 + 6x + 3$ shown below, we look for the point where the curve is parallel to the $x$ axis.

The coordinates of the turning point are approximately $(-1.5, -1.5)$.

Deducing Roots Algebraically

To find the roots of $f(x) = ax^2 + bx + c$ algebraically, we solve the equation $ax^2 + bx + c = 0$ using methods such as factorisation, the quadratic formula, or completing the square.

Example: Find the roots of $f(x) = 2x^2 + 5x + 3$. We solve $2x^2 + 5x + 3 = 0$. Factoring the expression gives $(2x + 3)(x + 1) = 0$. Therefore, $2x + 3 = 0$ or $x + 1 = 0$, leading to roots at $x = -1.5$ and $x = -1$.

Completing the Square

Completing the square is an algebraic process used to rewrite a quadratic expression of the form $x^2 + kx$ as a difference of two squares: $x^2 + kx = (x + \frac{k}{2})^2 - (\frac{k}{2})^2$

Worked Examples of Completing the Square:

1. For $x^2 + 4x$: Halve the coefficient of $x$ to get $2$. Then, $x^2 + 4x = (x + 2)^2 - 2^2$.
2. For $x^2 - 6x$: Half of $-6$ is $-3$. Thus, $x^2 - 6x = (x - 3)^2 - (-3)^2 = (x - 3)^2 - 3^2$.
3. For $2x^2 + 6x$: First, factor out the coefficient of $x^2$ so that the internal $x^2$ coefficient is 1: $2(x^2 + 3x)$. Now complete the square inside the bracket: $2[(x + \frac{3}{2})^2 - (\frac{3}{2})^2]$. Distributing the 2 gives $2(x + \frac{3}{2})^2 - \frac{9}{2}$.

Finding the Turning Point Algebraically

By completing the square, we can transform $y = ax^2 + bx + c$ into the form $y = a(x + p)^2 + q$. In this form, the turning point occurs when the squared term $(x + p)^2$ is at its minimum value, which is zero. This happens when $x = -p$, making the $y$ value equal to $q$. Thus, the turning point is $(-p, q)$.

Worked Example 1: Find the turning point of $y = x^2 + 6x + 7$. Completing the square: $y = (x + 3)^2 - 9 + 7 = (x + 3)^2 - 2$. The turning point occurs when $x + 3 = 0$, giving $x = -3$ and $y = -2$. The turning point is $(-3, -2)$.

Worked Example 2: Find the turning point of $y = 2x^2 + 6x + 5$. First, factor out the 2: $y = 2(x^2 + 3x) + 5$. Completing the square inside: $y = 2[(x + \frac{3}{2})^2 - (\frac{3}{2})^2] + 5$. Simplifying: $y = 2(x + \frac{3}{2})^2 - \frac{9}{2} + 5 = 2(x + \frac{3}{2})^2 + \frac{1}{2}$. The turning point occurs when $x + \frac{3}{2} = 0$, giving $x = -1.5$ and $y = 0.5$. The turning point is $(-1.5, 0.5)$.