Problem 70 - Elementary Algebra

Question

Determine the x- and y-intercepts. $$ y=-x 2+10 x-25 $$

Step-by-Step Solution

Answer

The y-intercept is (0, -25) and the x-intercept is (5, 0).

Step 1: Identify the Equation

We start with the given equation for a parabola: \( y = -x^2 + 10x - 25 \). The task is to find the x- and y-intercepts of this equation.

Step 2: Finding the Y-Intercept

To find the y-intercept, set \( x = 0 \) in the equation and solve for \( y \). This gives us:\[ y = -0^2 + 10 \cdot 0 - 25 = -25. \]The y-intercept is the point \((0, -25)\).

Step 3: Finding the X-Intercepts

To find the x-intercepts, set \( y = 0 \) and solve the equation \(-x^2 + 10x - 25 = 0\).

Step 4: Solving the Quadratic Equation

We use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -1 \), \( b = 10 \), \( c = -25 \).Calculate the discriminant: \[ \Delta = b^2 - 4ac = 10^2 - 4(-1)(-25) = 100 - 100 = 0. \]

Step 5: Finding the Roots

Since the discriminant is zero, there is exactly one real root. Using the quadratic formula:\[ x = \frac{-10 \pm \sqrt{0}}{2(-1)} = \frac{-10}{-2} = 5. \]Thus, the x-intercept is the point \((5, 0)\).

Key Concepts

Quadratic Equation

The quadratic equation is a fundamental concept in algebra and appears in the form \( ax^2 + bx + c = 0 \). This type of equation is called 'quadratic' because the highest exponent of the variable \( x \) is 2. It describes a parabola when plotted on a graph. A parabola is a U-shaped curve that can open upwards or downwards depending on the sign of the coefficient \( a \).

• If \( a > 0 \), the parabola opens upwards.
• If \( a < 0 \), it opens downwards, as in our example equation \( y = -x^2 + 10x - 25 \).

Every quadratic equation can have up to two unique solutions, or roots, which may be real or complex. These roots determine where the parabola intersects the x-axis, an important step in finding the x-intercepts.

X-Intercept

The x-intercepts of a graph are the points where the graph crosses the x-axis. In the context of a quadratic equation, we find these by setting \( y = 0 \). This step transforms our quadratic equation into \( -x^2 + 10x - 25 = 0 \) and involves solving for \( x \).Finding the x-intercepts means we're looking for the values of \( x \) that make the equation equal to zero, essentially finding the roots of the equation. For our specific example, we used the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]To calculate these, first, we identified the coefficients \( a = -1 \), \( b = 10 \), and \( c = -25 \) from the equation. The discriminant \( \Delta \) calculated as zero, implying the equation has exactly one real root, \( x = 5 \). Therefore, the x-intercept for this parabola is the point \((5, 0)\).

Y-Intercept

The y-intercept is a point where the graph crosses the y-axis, which happens when \( x = 0 \). To find it, we substitute \( x = 0 \) into the given quadratic equation and solve for \( y \). This is a straightforward calculation.For our equation \( y = -x^2 + 10x - 25 \), by setting \( x = 0 \), we get:\[ y = -(0)^2 + 10 \cdot 0 - 25 = -25 \]Thus, the y-intercept is at the point \((0, -25)\). This point is where the graph would intersect the y-axis, showing us the value when all other variables are zero. Understanding intercepts is important because they provide essential information about the behavior and position of a parabola on the coordinate plane.