Problem 26 - Elementary Algebra

Question

Graph. Find the vertex and the y-intercept. In addition, find the \(x\) - intercepts if they exist. $$ y=-x 2-2 x+15 $$

Step-by-Step Solution

Answer

Vertex: (-1, 16). Y-intercept: (0, 15). X-intercepts: (3, 0) and (-5, 0).

Step 1: Identify the Quadratic Equation Format

The given equation is in the standard quadratic form: \[ y = -x^2 - 2x + 15 \] This equation is written as \( y = ax^2 + bx + c \) , where \( a = -1 \), \( b = -2 \), and \( c = 15 \).

Step 2: Find the Vertex of the Parabola

The vertex of a parabola with the equation \( y = ax^2 + bx + c \) can be found using the formula: \[ x = -\frac{b}{2a} \] Plug in the values: \[ x = -\frac{-2}{2(-1)} = \frac{2}{-2} = -1 \] Now substitute \( x = -1 \) back into the equation to find \( y \): \[ y = -(-1)^2 - 2(-1) + 15 = -1 + 2 + 15 = 16 \] Thus, the vertex is \((-1, 16)\).

Step 3: Find the Y-intercept

The y-intercept of the equation is found by setting \( x = 0 \) and solving for \( y \). Using the equation: \[ y = -0^2 - 2(0) + 15 = 15 \] Therefore, the y-intercept is \((0, 15)\).

Step 4: Find the X-intercepts

X-intercepts occur where \( y = 0 \). Set the equation equal to zero: \[ 0 = -x^2 - 2x + 15 \] Rewriting gives: \[ x^2 + 2x - 15 = 0 \] This can be factored or solved using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] With \( a = 1, b = 2, c = -15 \), compute: \[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-15)}}{2(1)} \] \[ x = \frac{-2 \pm \sqrt{4 + 60}}{2} \] \[ x = \frac{-2 \pm \sqrt{64}}{2} \] \[ x = \frac{-2 \pm 8}{2} \] This gives \( x = 3 \) and \( x = -5 \). Therefore, the x-intercepts are \((3, 0)\) and \((-5, 0)\).

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a significant point that indicates the peak or the lowest point of the curve, depending on whether the parabola opens upwards or downwards. This unique point helps us understand the maximum or minimum value of the quadratic function.
To find the vertex, we use the formula for the x-coordinate:

• For the equation in the form of \( y = ax^2 + bx + c \), the x-coordinate of the vertex is calculated by \( x = -\frac{b}{2a} \).

After calculating the x-coordinate, substitute it back into the original equation to get the y-coordinate. This pair \((x, y)\) forms the vertex of the parabola. If the coefficient \(a\) is negative, like in our example \( y = -x^2 - 2x + 15 \), the parabola opens downward, and thus the vertex is the highest point.
The vertex for this equation is calculated to be \((-1, 16)\). This means that at \(x = -1\), the parabola reaches its highest point of \(y = 16\). Understanding this is crucial because the vertex tells us whether the function has a maximum or minimum value and exactly where it occurs.

Y-intercept

The y-intercept is the point where the parabola crosses the y-axis. It is found very simply by setting the value of \(x\) to zero in the quadratic equation. This is because every point along the y-axis has an x-coordinate of zero.
The significance of finding the y-intercept lies in understanding the initial value of the function when the input (x) is at its starting position of zero.

• Plug \( x = 0 \) into the equation: \( y = -0^2 - 2(0) + 15 \).
• This simplifies to \( y = 15 \).

Thus, the y-intercept is the point \((0, 15)\). This means when the input \(x\) is zero, the output \(y\) is 15. The y-intercept is a critical part of graphing the parabola, providing you an anchor point to start your graphical plot of the quadratic equation.

X-intercepts

X-intercepts are the points where the parabola crosses the x-axis. At these points, the function's output value, \(y\), is zero. Identifying the x-intercepts involves solving the equation by setting \(y = 0\).

• Start with the equation: \( 0 = -x^2 - 2x + 15 \).
• Solve for \(x\) using factoring or the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

For the given quadratic equation, when solved by using the quadratic formula, we find:

• \( x = \frac{-2 \pm \sqrt{64}}{2} \), leading to solutions \( x = 3 \) and \( x = -5 \).

Therefore, the x-intercepts are at the points \((3, 0)\) and \((-5, 0)\). These intercepts show where the parabola intersects the x-axis, providing points that help define the shape and position of the parabola on the coordinate plane. By determining these intercepts, we gain insights into the roots of the quadratic equation, showing where the function equals zero.