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The vertex of a parabola represents its highest or lowest point, depending on the graph's orientation. For the quadratic function in the standard form, \( f(x) = ax^2 + bx + c \), the vertex can be calculated using the formula:

• \( x = -\frac{b}{2a} \)

This equation gives us the x-coordinate of the vertex. By substituting this x-value back into the function, we find the y-coordinate. Let's break down the vertex calculation for the given quadratic function:

• Given function: \( f(x) = x^2 + 10x + 15 \)
• Coefficients: \( a = 1, b = 10, c = 15 \)

First, calculate the x-coordinate of the vertex:

• \( x = -\frac{10}{2(1)} = -5 \)

Next, find the y-coordinate by substituting \( x = -5 \) back into the function:

• \( f(-5) = (-5)^2 + 10(-5) + 15 = 25 - 50 + 15 = -10 \)

So, the vertex of the parabola is \((-5, -10)\). This point will either be the minimum or maximum of the function depending on whether the parabola opens upwards or downwards.

The y-intercept of a quadratic function is the point where the graph intersects the y-axis. To find this, simply evaluate the quadratic function at \( x = 0 \). The y-intercept will be given by the constant term \( c \). For our function:

• Function: \( f(x) = x^2 + 10x + 15 \)
• Y-intercept Calculation: \( f(0) = 0^2 + 10(0) + 15 = 15 \)

Thus, the y-intercept is \((0, 15)\). This tells us that, when you substitute \( x = 0 \) into the function, the output is \( 15 \). It confirms where the graph will cross the y-axis, providing a starting point for sketching the curve.

Finding the x-intercepts of a quadratic function means solving for the points where the graph crosses the x-axis. This occurs when the function \( f(x) = 0 \). To find the x-intercepts, we solve the equation using the quadratic formula:

• Quadratic Formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Let’s solve for the x-intercepts of our function \( f(x) = x^2 + 10x + 15 \):

• Coefficients: \( a = 1, b = 10, c = 15 \)
• Discriminant: \( b^2 - 4ac = 10^2 - 4 \times 1 \times 15 = 100 - 60 = 40 \)
• X-intercepts Computation: \( x = \frac{-10 \pm \sqrt{40}}{2} \)
• Solutions: \( x \approx -1.8 \) and \( x \approx -8.2 \)

These approximate solutions \((-1.8, 0)\) and \((-8.2, 0)\) are the points where the parabola crosses the x-axis, which are crucial for graphing the parabola.

Graphing a quadratic function involves plotting key points including the vertex, y-intercept, and x-intercepts. It is important to understand the graph's orientation:

• Direction: The parabola opens upwards if the coefficient \( a > 0 \) and downwards if \( a < 0 \).

Given our function \( f(x) = x^2 + 10x + 15 \), the coefficient \( a = 1 \) tells us the parabola opens upwards. Here's how you can sketch it:

• Plot the vertex at \((-5, -10)\).
• Mark the y-intercept at \((0, 15)\).
• Locate the x-intercepts approximately at \((-1.8, 0)\) and \((-8.2, 0)\).

Once these key points are on the graph, draw a smooth curve through them, representing the parabola. This visual representation helps in understanding how the function behaves across the x-y plane, highlighting the symmetry and key intersections of the quadratic equation.