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Graphing quadratic functions is a pivotal skill in algebra, allowing us to visualize these functions with their characteristic "U"-shaped curves, called parabolas. The general form of a quadratic function is expressed as \[ f(x) = ax^2 + bx + c \] where \(a\), \(b\), and \(c\) are constants. The graph either opens upwards (\(a > 0\)) or downwards (\(a < 0\)).
For our quadratic function \(f(x) = -3x^2 + 5x + 4\), the graph will open downward, as the leading coefficient \(a = -3\) is negative. To sketch the graph:

• Locate the vertex, which provides a starting point.
• Determine the y-intercept by evaluating \(f(0)\). This gives the initial data point for the graph.
• Find the x-intercepts using the quadratic formula, which shows where the graph crosses the x-axis.
• Plot these points, and draw a smooth curve through them, forming the parabola.

Understanding this process helps in visualizing solutions and analyzing the function's behavior.

Intercepts are the points where the quadratic function crosses the axes. Calculating these provides a deeper understanding of the function's graph.

**Y-intercept**To find the y-intercept, set \(x = 0\). For \(f(x) = -3x^2 + 5x + 4\), this results in \(f(0) = 4\). So, the y-intercept is the point \((0, 4)\), indicating where the graph intersects the y-axis.

**X-intercepts**To find the x-intercepts, set the function equal to zero and solve the resulting equation:\[-3x^2 + 5x + 4 = 0\]Using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Substitute the values \(a = -3\), \(b = 5\), and \(c = 4\) to find the roots:

• \(x_1 = -\frac{1}{3}\)
• \(x_2 = 4\)

These points, \((-\frac{1}{3}, 0)\) and \((4, 0)\), represent where the graph crosses the x-axis, crucial for improving the graph's accuracy.

Quadratic functions have a domain of all real numbers, \(( -\infty, \infty )\), because they can take any real number as input. However, the range is tied to the vertical positioning and direction in which the parabola opens.

**Domain**Since the function \(f(x) = -3x^2 + 5x + 4\) is a quadratic, it has no restrictions on its x-values, hence domain is all real numbers.

**Range**The vertex of the quadratic function, \(\left(\frac{5}{6}, \frac{22}{3}\right)\), is the highest point due to the negative leading coefficient (downward-opening parabola). Therefore, the range is negative infinity to the vertex's y-component:\[(-\infty, \frac{22}{3}] \].
This provides full coverage of potential output values for the function.

The vertex of a quadratic function provides critical information about its graph. It's either the lowest or highest point on the parabola, depending on whether it opens upwards or downwards.

For a function in general form like \(f(x) = -3x^2 + 5x + 4\), the vertex coordinates \((h, k)\) can be found using:

• \(h = -\frac{b}{2a} = -\frac{5}{2(-3)} = \frac{5}{6}\)
• Substitute \(h\) in the function to find \(k\):\[k = f\left(\frac{5}{6}\right) = \frac{22}{3} \]

The vertex, \(\left( \frac{5}{6}, \frac{22}{3} \right)\), signifies the peak maximum point due to the downward opening. Knowing the vertex helps in graphing and understanding where the function changes direction.

The axis of symmetry is an essential feature of a quadratic graph. It's a vertical line that passes through the vertex, dividing the parabola into two mirror-image halves.
For the function \( f(x) = -3x^2 + 5x + 4 \), the equation of the axis of symmetry is \[ x = \frac{5}{6} \].
Since it passes through the vertex \(\left(\frac{5}{6}, \frac{22}{3}\right)\), it directly relates to the formula:\[ x = -\frac{b}{2a} \]Every point on the parabola has a corresponding point on the opposite side of this line, precisely equidistant from it.
This property is useful when determining the function's minimum or maximum and aids in graphing.