Problem 19 - Precalculus

Question

Determine any \(x\) - or \(y\) -intercepts for the graph of the equation. Note: You're not asked to draw the graph. (a) \(y=x^{2}+x-1\) (b) \(y=x^{2}+x+1\)

Step-by-Step Solution

Answer

(a) x-intercepts: \(\frac{-1\pm\sqrt{5}}{2}\), y-intercept: \(-1\); (b) no x-intercepts, y-intercept: \(1\).

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

To find the x-intercepts, set \(y = 0\) in the equation \(y = x^2 + x - 1\) and solve for \(x\). This gives us the quadratic equation:\[ x^2 + x - 1 = 0 \]Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = 1, c = -1\), we find the roots:\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]\[ x = \frac{-1 \pm \sqrt{1 + 4}}{2} \]\[ x = \frac{-1 \pm \sqrt{5}}{2} \]Therefore, the x-intercepts are \(x = \frac{-1 + \sqrt{5}}{2}\) and \(x = \frac{-1 - \sqrt{5}}{2}\).

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

The y-intercept occurs when \(x = 0\). Substitute \(x = 0\) into the equation \(y = x^2 + x - 1\):\[ y = 0^2 + 0 - 1 \]\[ y = -1 \]Hence, the y-intercept for equation (a) is \(y = -1\).

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

Set \(y = 0\) in the equation \(y = x^2 + x + 1\) and solve for \(x\):\[ x^2 + x + 1 = 0 \]Using the quadratic formula with \(a = 1, b = 1, c = 1\), we calculate:\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]\[ x = \frac{-1 \pm \sqrt{1 - 4}}{2} \]\[ x = \frac{-1 \pm \sqrt{-3}}{2} \]Since the discriminant \(1 - 4\) is negative, there are no real roots, meaning there are no x-intercepts for equation (b).

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

The y-intercept occurs when \(x = 0\). Substitute \(x = 0\) into the equation \(y = x^2 + x + 1\):\[ y = 0^2 + 0 + 1 \]\[ y = 1 \]Hence, the y-intercept for equation (b) is \(y = 1\).

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions of the form \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. These equations describe parabolas, which can open upwards or downwards depending on the sign of the leading coefficient \(a\). The solutions to quadratic equations are found using various methods:

• Factoring: Sometimes quadratic equations can be factored into simpler expressions. However, not all quadratics are factorable using integers.
• Completing the square: This involves manipulating the equation to see it as a perfect square trinomial.
• Quadratic formula: A universal method expressed as \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). It can solve any quadratic equation, given real or complex roots.

The discriminant, \(b^2 - 4ac\), determines the nature of the roots:

• If the discriminant is positive, the equation has two real and distinct roots.
• If it is zero, there is exactly one real root (or a repeated root).
• If the discriminant is negative, the roots are complex and the equation has no real solution.

X-Intercepts

The x-intercepts of a graph are the points where the graph crosses the x-axis. At these points, the value of \(y\) is zero. For quadratic equations, finding the x-intercepts involves setting the equation equal to zero and solving for \(x\). This gives us the roots of the quadratic equation.
In our examples, we solve two quadratics:

• For equation (a) \(y = x^2 + x - 1\), the x-intercepts are found using the quadratic formula. With a positive discriminant \(b^2 - 4ac = 5\), it yields two real roots: \(x = \frac{-1 + \sqrt{5}}{2}\) and \(x = \frac{-1 - \sqrt{5}}{2}\).
• For equation (b) \(y = x^2 + x + 1\), the discriminant is negative, indicating no real roots. Hence, there are no x-intercepts for this graph.

By analyzing the discriminant, we can determine whether the graph will cross the x-axis, and where, if applicable.

Y-Intercepts

Y-intercepts are where the graph crosses the y-axis. This occurs when \(x = 0\). To find the y-intercept of a quadratic equation, substitute \(x = 0\) in the equation and solve for \(y\).
In our examples:

• For equation (a) \(y = x^2 + x - 1\), substituting \(x = 0\) gives \(y = -1\). Therefore, the y-intercept is \(y = -1\).
• For equation (b) \(y = x^2 + x + 1\), substituting \(x = 0\) results in \(y = 1\). Thus, the y-intercept is \(y = 1\).

The y-intercept provides a point on the graph where it touches or crosses the y-axis. This simple substitution offers an easy means to graphically validate that the function behaves as expected.