Problem 99 - Algebra 1

Question

Sketch the graph of the function. Label the vertex. y=-2 x^{2}-3 x+2

Step-by-Step Solution

Answer

The sketch of the function \(y=-2 x^{2}-3 x+2\) is a parabola opening downward, with its vertex at \((3/4, -2.125)\) and Y-intercept at \((0, 2)\).

Step 1: Find the vertex

The vertex of a quadratic function in the form \(y = a x^{2} + b x + c\) is given by the point \((-b/(2a), f(-b/(2a))\). Substituting the values \(a=-2\) and \(b=-3\), the x-coordinate of the vertex is given by \(-(-3)/(2*(-2)) = 3/4\). Substituting \(x=3/4\) into the function, the y-coordinate of the vertex is \(-2*(3/4)^{2} - 3*(3/4) + 2 = -2.125\). Hence, the vertex of the function is \((3/4, -2.125)\).

Step 2: Find the Y-intercept

The Y-intercept of the function is the value of the function at \(x = 0\). Substituting \(x=0\) into the function gives \(y = -2*(0)^{2} - 3*0 + 2 = 2\). Hence, the Y-intercept of the function is at the point \((0, 2)\).

Step 3: Sketch the graph

Now sketch the graph. Start by plotting the vertex \((3/4, -2.125)\) and the Y-intercept \((0,2)\). Since the coefficient of \(x^{2}\) is negative, the graph opens downward. Draw a graph that starts at the Y-intercept, reaches a maximum at the vertex, and then continues downward. Label the vertex.

Key Concepts

Vertex

The vertex of a quadratic function serves as a key feature that helps us understand the shape and position of its graph. In the quadratic function \[y = -2x^2 - 3x + 2\], the vertex can be found using the formula for the vertex of a parabola, \((-\frac{b}{2a}, f(-\frac{b}{2a}))\). Here, \(a\), \(b\), and \(c\) are the coefficients in the standard form. For our given function, \(a = -2\) and \(b = -3\).

• First, calculate the x-coordinate of the vertex: \[\frac{-(-3)}{2(-2)} = \frac{3}{4}\]
• Substitute \(x = \frac{3}{4}\) into the function to find the y-coordinate: \[y = -2\left(\frac{3}{4}\right)^2 - 3\left(\frac{3}{4}\right) + 2 = -2.125\]

Thus, the vertex of the function is \((\frac{3}{4}, -2.125)\). This point is where the parabola changes direction and provides a maximum or minimum value depending on whether the parabola opens upward or downward.

Y-intercept

To find the Y-intercept of a quadratic function, you simply set \(x = 0\) in the equation and solve for \(y\). This gives us the point at which the graph crosses the Y-axis. For the function \[y = -2x^2 - 3x + 2\], calculating the Y-intercept is straightforward.

• Substitute \(x = 0\) into the equation: \[y = -2(0)^2 - 3(0) + 2 = 2\]

The Y-intercept is at point \((0, 2)\). This is where the graph of the parabola will touch the Y-axis. It acts as a starting point or one of the reference points used to sketch the graph. A clear understanding of the Y-intercept aids in visualizing the initial position of the parabola relative to the axes.

Parabola Opens Downward

The direction in which a parabola opens is determined by the coefficient of \(x^2\) in the quadratic equation. Here, the function \[y = -2x^2 - 3x + 2\] has a negative coefficient \(a = -2\), which means the parabola opens downward.

• Negative coefficient \(a\): Parabola opens downward - If \(a < 0\), like in our function, the parabola depends to open downwards forming an inverted "U" shape.
• Positive coefficient \(a\): Parabola opens upward - If \(a > 0\), the parabola is an upright "U" shape.

When a parabola opens downward, the vertex represents the maximum point on the graph. This is crucial for understanding the function's behavior as it helps to identify the peak value that the quadratic function can reach. In real-world applications, recognizing this shape can be critical in situations involving maximum values, such as profit or altitude in trajectory problems.