Question

Find an equation of the parabola that has the indicated vertex and whose graph passes through the given point. Vertex: \((5,12) ;\) point: \((7,15)\)

Short answer

The equation of the parabola that passes through the vertex (5,12) and the point (7,15) is \(y=(3/4)(x-5)^2+12 \).

Step-by-step solution

Substituting the vertex into the equation.

Replace \(h\) with \(5\) and \(k\) with \(12\) in the vertex form of a parabola \(y=a(x-h)^2+k\). This results in: \(y=a(x-5)^2+12 \)

Substitute the given point into the formula.

Now, substitute the given point (7,15) into the equation: \(15=a(7-5)^2 + 12\). Hence, \(15=a(2)^2 +12\). So, \(15=4a+12\).

Solve for 'a'.

The equation \(15=4a+12\) simplifies to \(4a = 15 - 12\) and \(4a = 3\). Divide by 4 to solve for 'a', yielding \(a = 3/4\).

Key concepts

Vertex Form of a Parabola

Understanding the vertex form of a parabola is crucial for graphing and analyzing the properties of parabolic equations. The vertex form is given by the equation \(y = a(x-h)^2 + k\), where \(h\) and \(k\) represent the coordinates of the parabola's vertex, and \(a\) is a coefficient that determines the width and direction of the parabola. When \(a > 0\), the parabola opens upwards, and when \(a < 0\), it opens downwards. The greater the absolute value of \(a\), the narrower the parabola is.

For the exercise provided, the vertex is given as \(5, 12\). Plugging these values into the vertex form, we get \(y = a(x - 5)^2 + 12\). This rewritten equation represents a parabola that has its vertex at point \(5, 12\) and is pivotal for solving the given problem. The vertex form makes it easier to visualize the graph and also simplifies the process of finding the missing coefficient \(a\) once an additional point the parabola passes through is known.

Solving for Coefficients

When tasked with finding the equation of a parabola, one must often solve for unknown coefficients. The coefficient \(a\) in the vertex form \(y = a(x-h)^2 + k\) impacts the parabola's opening and its width. To determine \(a\), we use a given point through which the parabola passes.

In this exercise, after substituting the vertex into the vertex form, we plugged in the coordinates of the additional point \(7, 15\). By substituting these values into the equation \(15 = a(7 - 5)^2 + 12\), we set up an equation where \(a\) is the only unknown. Solving this equation for \(a\), by isolating it on one side, gave us the value of \(a = 3/4\). This process of solving for \(a\) is an application of basic algebraic principles, specifically the method of isolating the variable to find its value.

Quadratic Functions

Quadratic functions are polynomial functions of degree two, typically written in the standard form \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants and \(a \eq 0\). They graph as parabolas and have various characteristics such as a vertex, axis of symmetry, and potentially real or complex roots depending on the discriminant \(b^2 - 4ac\).

The equation from our exercise, once we found \(a = 3/4\), can be rewritten in the standard form by expanding the binomial and simplifying. However, working with the vertex form is often more intuitive when given the vertex and a point, as we can visualize the parabola's shape and position without additional steps. Vertex form also aids in identifying the parabola's maximum or minimum value, which occurs at the vertex, hence why it was useful for this problem.