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Understanding vertex calculation is crucial when dealing with quadratic functions in the form of \( f(x) = ax^2 + bx + c \). The vertex is a significant point since it represents the minimum or maximum of the parabola, depending on whether it opens upwards or downwards. To find the vertex, use the formula \( x = -\frac{b}{2a} \) to determine the x-coordinate.

In our function \( f(x) = 2x^2 + 4x + 3 \), \( a = 2 \) and \( b = 4 \). Substitute these values into the formula:

• \( x = -\frac{4}{2 \times 2} = -1 \)

Now, plug \( x = -1 \) back into the function to find the y-coordinate of the vertex:

• \( f(-1) = 2(-1)^2 + 4(-1) + 3 = 2 - 4 + 3 = 1 \)

Therefore, the vertex of the parabola is \((-1, 1)\). This point indicates both the lowest point on the graph because \( a > 0 \), reflecting the parabola's upward opening.

X-intercepts are the points where the graph of the function crosses the x-axis. For those points, the function value is zero, so we need to solve \( f(x) = 0 \).

Given the quadratic function \( f(x) = 2x^2 + 4x + 3 \), set it to zero:

• \( 2x^2 + 4x + 3 = 0 \)

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve.

• \( a = 2 \), \( b = 4 \), \( c = 3 \)
• Calculate the discriminant: \( b^2 - 4ac = 16 - 24 = -8 \)

Since the discriminant is negative, \( \sqrt{-8} \) is complex, meaning there are no real solutions and thus no x-intercepts. The graph does not cross the x-axis, indicating that the entire parabola is located above the x-axis.

Calculating the y-intercept of a quadratic function is straightforward. It is the point where the graph intersects the y-axis, which occurs when \( x = 0 \). To find this point, substitute \( x = 0 \) into the quadratic function.

• For \( f(x) = 2x^2 + 4x + 3 \), substitute to get \( f(0) = 3 \)

Therefore, the y-intercept is the point \((0, 3)\). This point represents the height where the parabola initially intersects the y-axis, giving us a point of reference on the vertical axis when plotting the graph.

Sketching the graph of a quadratic function involves plotting its critical points and establishing the curve's shape. Here's how you can do it:First, plot the vertex and the y-intercept on the coordinate plane.

• Vertex: \((-1, 1)\)
• Y-intercept: \((0, 3)\)

The absence of x-intercepts tells us the parabola does not cross the x-axis.

Since the leading coefficient \( a = 2 \) is positive, the parabola opens upwards. This shapes the curve above the x-axis, with the vertex at its lowest point.

Draw a smooth curve starting from the vertex, passing through the y-intercept, and extending upwards symmetrically on both sides. Make sure the graph depicts a U-shape, typical of quadratic functions with positive \( a \). This visualization helps understand how the function behaves and confirms characteristics like vertex position and intercepts.