Problem 18 - Precalculus

Question

Determine any \(x\) - or \(y\) -intercepts for the graph of the equation. Note: You're not asked to draw the graph. (a) \(y=x^{2}-4 x-12\) (b) \(y=x^{2}-4 x+12\)

Step-by-Step Solution

Answer

(a) x-intercepts at x = 6, x = -2; y-intercept at y = -12. (b) No real x-intercepts; y-intercept at y = 12.

Step 1: Determine the x-intercepts for equation (a)

To find the x-intercepts of the equation \( y = x^2 - 4x - 12 \), set \( y = 0 \) and solve for \( x \). This gives the equation \( x^2 - 4x - 12 = 0 \). Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -4 \), and \( c = -12 \).First, calculate the discriminant: \( b^2 - 4ac = (-4)^2 - 4(1)(-12) = 16 + 48 = 64 \).Now, apply the quadratic formula: \( x = \frac{-(-4) \pm \sqrt{64}}{2 \times 1} = \frac{4 \pm 8}{2} \).This results in two solutions: \( x = 6 \) and \( x = -2 \). These are the x-intercepts.

Step 2: Determine the y-intercept for equation (a)

The y-intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the equation \( y = x^2 - 4x - 12 \) to find the y-intercept:\( y = 0^2 - 4(0) - 12 = -12 \).Thus, the y-intercept is \( y = -12 \).

Step 3: Determine the x-intercepts for equation (b)

To find the x-intercepts of the equation \( y = x^2 - 4x + 12 \), set \( y = 0 \) and solve for \( x \). This gives the equation \( x^2 - 4x + 12 = 0 \). Use the quadratic formula where \( a = 1 \), \( b = -4 \), and \( c = 12 \).Calculate the discriminant: \( b^2 - 4ac = (-4)^2 - 4(1)(12) = 16 - 48 = -32 \).Since the discriminant is negative, there are no real x-intercepts.

Step 4: Determine the y-intercept for equation (b)

The y-intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the equation \( y = x^2 - 4x + 12 \) to find the y-intercept:\( y = 0^2 - 4(0) + 12 = 12 \).Thus, the y-intercept is \( y = 12 \).

Key Concepts

Quadratic Equations

A quadratic equation is a polynomial of degree two that can be written in the form \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants and \( a \) is not zero. These equations form a parabola when graphed on a coordinate plane. The shape and direction of the parabola depend on the value of \( a \):

• If \( a > 0 \), the parabola opens upward.
• If \( a < 0 \), the parabola opens downward.

The key features of quadratic equations include the vertex, axis of symmetry, and intercepts. Solving a quadratic equation involves finding the values of \( x \) (the roots) that make the equation equal to zero. This can be achieved via factoring, completing the square, or using the quadratic formula.

x-Intercepts

The x-intercepts (or roots) of a quadratic equation are the points where the graph crosses the x-axis. To find the x-intercepts, you set \( y = 0 \) in the equation and solve for \( x \). This results in the quadratic equation \( ax^2 + bx + c = 0 \). Solving this equation will provide the x-intercepts if they exist.
For example, in the case of the equation \( y = x^2 - 4x - 12 \), substituting \( y = 0 \) leads to \( x^2 - 4x - 12 = 0 \). Using the quadratic formula, two real solutions \( x = 6 \) and \( x = -2 \) are found, indicating two x-intercepts.
In contrast, for \( y = x^2 - 4x + 12 \), the discriminant is negative, revealing that there are no real x-intercepts for this equation because the parabola does not touch the x-axis.

y-Intercepts

The y-intercept of any equation is the point where the graph crosses the y-axis. This occurs when \( x = 0 \). To find the y-intercept of a quadratic equation, substitute \( x = 0 \) into the equation and solve for \( y \).
Take the equation \( y = x^2 - 4x - 12 \) as an example: by setting \( x = 0 \), you find \( y = -12 \). Thus, the y-intercept is \( y = -12 \).
Similarly, for \( y = x^2 - 4x + 12 \), when \( x = 0 \), \( y = 12 \). Hence, the graph of this equation intercepts the y-axis at \( y = 12 \). The y-intercept gives a quick snapshot of where the graph begins on the y-axis if the parabola were fully drawn.

Discriminant

The discriminant is a crucial component in solving quadratic equations and is derived from the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). The discriminant is represented by \( \Delta = b^2 - 4ac \).

• If \( \Delta > 0 \), there are two distinct real roots (the quadratic equation has two x-intercepts).
• If \( \Delta = 0 \), there is exactly one real root (the parabola touches the x-axis at one point).
• If \( \Delta < 0 \), there are no real roots (no x-intercepts, as the parabola does not touch the x-axis).

In our exercises, for \( y = x^2 - 4x - 12 \), the discriminant \( \Delta = 64 \), indicating two real roots. However, for \( y = x^2 - 4x + 12 \), the discriminant \( \Delta = -32 \), pointing to no real roots. Thus, the discriminant helps predict the nature of the roots of quadratic equations.