Problem 52 - Beginning and Intermediate Algebra

Question

Graph each equation using the vertex formula. Find the \(x\) - and \(y\) -intercepts. $$x=-\frac{3}{4} y^{2}+\frac{3}{2} y-\frac{11}{4}$$

Step-by-Step Solution

Answer

The vertex of the given quadratic function is \((1, \frac{5}{2})\). The x-intercept is located at \(\left(-\frac{11}{4},0\right)\). There are no real y-intercepts, as the discriminant in the quadratic formula is negative.

Step 1: Find the vertex

To find the vertex, we first need to find the coordinates of the vertex \((h, k)\). Since the given equation is a quadratic function in terms of \(y\), we can find the vertex by using the vertex formula: $$ h = \frac{-b}{2a} $$ where, in our equation, $$ a = -\frac{3}{4}, b = \frac{3}{2} $$ Now, substitute the values of \(a\) and \(b\) into the vertex formula: $$ h = \frac{-\frac{3}{2}}{2 \cdot -\frac{3}{4}} = \frac{3}{3} = 1 $$ Once we find the \(h\)-coordinate, we can substitute this value back into the original equation to find the \(k\)-coordinate: $$ k = -\frac{3}{4}(1)^2 + \frac{3}{2}(1) - \frac{11}{4} = -\frac{3}{4} + \frac{3}{2} - \frac{11}{4}= \frac{5}{2} $$ So, the vertex is \((1,\dfrac{5}{2})\).

Step 2: Find the x-intercept

To find the x-intercept, we need to set \(y=0\) in the equation and solve for \(x\). Substituting \(y=0\): $$ x = -\frac{3}{4}(0)^2 + \frac{3}{2}(0) - \frac{11}{4} = -\frac{11}{4} $$ The x-intercept is located at \(\left(-\frac{11}{4},0\right)\).

Step 3: Find the y-intercept

To find the y-intercept, we need to set \(x=0\) and solve for \(y\). Substituting \(x=0\): $$ 0=-\frac{3}{4} y^{2}+\frac{3}{2} y-\frac{11}{4} $$ Now we need to solve this quadratic equation for \(y\). We can multiply the equation by \(4\) to clear the denominators: $$ 0 = -3y^2 + 6y - 11 $$ This quadratic equation does not factor easily, so we use the quadratic formula to solve for \(y\): $$ y = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} $$ Where \(A = -3\), \(B = 6\), and \(C = -11\): $$ y = \frac{-6 \pm \sqrt{(-6)^2 - 4(-3)(-11)}}{2(-3)} $$ Simplifying the expression: $$ y = \frac{-6 \pm \sqrt{36-132}}{-6} = \frac{-6 \pm \sqrt{-96}}{-6} $$ Here we can see that the discriminant (\(B^2 - 4AC = -96\)) is negative, so there are no real y-intercepts.

Key Concepts

quadratic function

A quadratic function is a type of polynomial that has a degree of 2. The general form of a quadratic function is given by the equation:

• \( ax^2 + bx + c = 0 \)

where \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). In this form, \( a \) determines the direction and width of the parabola, \( b \) affects the symmetry, and \( c \) indicates the y-intercept. Quadratic functions graph as parabolas, which are symmetrical curves that either open upwards or downwards.
In the given exercise, the quadratic function is presented in terms of \( y \), yielding x-values as outputs. Hence, the function is \( x = -\frac{3}{4}y^2 + \frac{3}{2}y - \frac{11}{4} \), where the corresponding parabola opens in a horizontal direction because the quadratic term is dependent on \( y \).

x-intercepts

The x-intercept of a function is the point where the graph crosses the x-axis. This occurs when the output, \( y \), is equal to zero. To find the x-intercept, we set \( y = 0 \) in the original equation and solve for \( x \).
For the exercise's quadratic function, applying \( y = 0 \) gives:

• \( x = -\frac{3}{4}(0)^2 + \frac{3}{2}(0) - \frac{11}{4} = -\frac{11}{4} \)

This indicates the x-intercept is at the point \( \left(-\frac{11}{4}, 0\right) \). Since the graph crosses the x-axis once, at this point, it reflects that there is only one x-intercept in this case. The negative value signifies that the intercept lies on the negative side of the x-axis.

y-intercepts

The y-intercept is where the graph crosses the y-axis, which is found by setting \( x = 0 \) in the equation and solving for \( y \). This task involves finding the values of \( y \) for which the equation \( x = -\frac{3}{4}y^2 + \frac{3}{2}y - \frac{11}{4} = 0 \) holds true.
With this particular quadratic expression:

• First, convert it to a more manageable format by multiplying through by 4 to eliminate denominators: \( 0 = -3y^2 + 6y - 11 \).

To determine if a quadratic expression has y-intercepts, check the discriminant \( B^2 - 4AC \). Here, it becomes clear:

• \( (-6)^2 - 4(-3)(-11) = 36 - 132 = -96 \)

A negative discriminant tells us there are no real y-intercepts, since the equation under the square root is negative, leading to no intersection with the y-axis.

quadratic formula

Solving quadratic equations often involves using the quadratic formula, particularly when other methods such as factoring or completing the square are not convenient. The quadratic formula provides the roots of any quadratic equation \( ax^2 + bx + c = 0 \):

• \( y = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \)

This formula is applicable to finding x- or y-values, depending on which variable represents the quadratic term. In the context of this exercise, using the quadratic formula helps determine potential y-intercepts. Given:

• \( A = -3 \), \( B = 6 \), and \( C = -11 \)

Substituting into the formula results in:

• \( y = \frac{-6 \pm \sqrt{36 - 132}}{-6} = \frac{-6 \pm \sqrt{-96}}{-6} \)

A negative value within the square root indicates no real solutions, confirming why the equation has no real y-intercepts. The quadratic formula is invaluable, especially in cases where the quadratic does not factorize easily.