Problem 76 - Elementary Algebra

Question

Graph. Find the vertex and the y-intercept. In addition, find the x-intercepts if they exist. $$ y=-x 2-6 x+7 $$

Step-by-Step Solution

Answer

Vertex: (-3, 16), y-intercept: (0, 7), x-intercepts: (-7, 0) and (1, 0).

Step 1: Identify the Quadratic Equation Format

In order to find key features of the graph of the equation, let's first identify the format of the quadratic equation. The given equation is:\[ y = -x^2 - 6x + 7 \]This can be rewritten in the standard form of a quadratic equation, which is:\[ y = ax^2 + bx + c \]Where \( a = -1 \), \( b = -6 \), and \( c = 7 \).

Step 2: Find the Vertex

The vertex form of a quadratic equation is \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex. To find \(h\), we use the formula:\[ h = -\frac{b}{2a} \]Plugging in the values:\[ h = -\frac{-6}{2(-1)} = -3 \]To find \(k\), substitute \(h\) back into the original equation:\[ k = -(-3)^2 - 6(-3) + 7 = -9 + 18 + 7 = 16 \]Thus, the vertex is \((-3, 16)\).

Step 3: Find the Y-Intercept

The y-intercept is found by substituting \( x = 0 \) into the original equation:\[ y = -0^2 - 6(0) + 7 = 7 \]Therefore, the y-intercept is \( (0, 7) \).

Step 4: Find the X-Intercepts

To find the x-intercepts, we set \( y = 0 \) and solve for \( x \):\[ 0 = -x^2 - 6x + 7 \]Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -1 \), \( b = -6 \), and \( c = 7 \):\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(-1)(7)}}{2(-1)} \]\[ x = \frac{6 \pm \sqrt{36 + 28}}{-2} \]\[ x = \frac{6 \pm \sqrt{64}}{-2} \]\[ x = \frac{6 \pm 8}{-2} \]Calculating the roots:\[ x_1 = \frac{6 + 8}{-2} = -7 \]\[ x_2 = \frac{6 - 8}{-2} = 1 \]The x-intercepts are \((-7, 0)\) and \((1, 0)\).

Key Concepts

Vertex

In a quadratic equation, the vertex is a crucial point as it represents the maximum or minimum point of the parabola. In our equation, \( y = -x^2 - 6x + 7 \), the vertex gives us the topmost point. To find the vertex, we employ the vertex formula. This formula is derived from rewriting the equation in vertex form, \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex.

To uncover \(h\), you use the handy formula \( h = -\frac{b}{2a} \). Plugging in our values, \( b = -6 \) and \( a = -1 \), results in:

• \( h = -\frac{-6}{2 \times -1} = -3 \)

To find \(k\), substitute \(h\) back into the equation:

• \( k = -(-3)^2 - 6(-3) + 7 = 16 \)

Thus, the vertex is located at \((-3, 16)\). This means the parabola peaks at \(x = -3\) and reaches a height of \(y = 16\). Knowing this helps you understand the curve's shape and position on the graph.

Y-Intercept

The y-intercept is where the graph intersects the y-axis, providing a starting point when plotting the graph of a quadratic equation. This point shows where the parabola crosses the line \( x = 0 \). To find this helpful intercept, you simply substitute \( x = 0 \) into the equation.

• For our equation, \( y = -0^2 - 6(0) + 7 \) simplifies to \( y = 7 \).

Therefore, the y-intercept of the quadratic \( y = -x^2 - 6x + 7 \) is \((0, 7)\). This point illustrates where the parabola cuts through the vertical axis. It also helps when sketching the graph because it gives a concrete point to start plotting the shape of the parabola.

X-Intercepts

X-intercepts reveal where the curve crosses the x-axis, essentially points where the value of \(y\) is zero. Solving for these intercepts helps to understand the parabola's width and location along the x-axis. To find the intercepts in \( y = -x^2 - 6x + 7 \), we set \( y = 0 \) and solve for \( x \) using the quadratic formula.

• The formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

With \( a = -1 \), \( b = -6 \), and \( c = 7 \), substitute these into the quadratic formula:

• \( x = \frac{6 \pm \sqrt{64}}{-2} \)
• The solutions are \( x_1 = -7 \) and \( x_2 = 1 \).

So, the x-intercepts are \((-7, 0)\) and \((1, 0)\). These points mark where the parabola intersects the x-axis, which is invaluable for understanding the graph's overall direction and span.