Question

passes through \((0,-24)\) and has vertex \((-6,-12)\)

Short answer

The equation of the parabola that passes through \((0,-24)\) and has a vertex at \((-6,-12)\) is \(y = -0.5(x+6)^2 - 12\).

Step-by-step solution

Identification of the vertex

The vertex of the parabola is given as \((-6,-12)\). Hence, in the vertex form \(y = a(x-h)^2 + k\), the values of 'h' and 'k' are -6 and -12 respectively. So, the partial equation is \(y = a(x+6)^2 - 12\)

Substitution of the point into the equation

Substitute the coordinates of the given point \((0,-24)\) into partial equation. This gives \(-24 = a(0+6)^2 - 12\).

Solve for 'a' coefficient

Solving the equation from step 2 for 'a' gives \(a = \frac{-24 + 12}{6^2} = -0.5\)

Write the Final Equation

Substitute the value of 'a' as found in step 3 into the partial equation from step 1. This gives the final equation for the parabola as \(y = -0.5(x+6)^2 - 12\).

Key concepts

Parabola Equation

The equation of a parabola can often be expressed in the vertex form, which is given by: \[ y = a(x - h)^2 + k \] where \((h, k)\) represents the vertex of the parabola and \(a\) is a coefficient that affects the graph's direction and width. If \(a\) is positive, the parabola opens upwards, while a negative \(a\) indicates it opens downwards.When given a vertex and a point that the parabola passes through, we can use these to determine the complete parabola equation. For instance, in our example, the vertex is \((-6, -12)\), so \(h = -6\) and \(k = -12\). This partially completes our equation: \[ y = a(x + 6)^2 - 12 \]By substituting a given point like \((0, -24)\) into this equation, we solve for \(a\) to find the full equation.

Solving Quadratic Equations

Quadratic equations are one of the critical components in algebra, with various forms and methods for solving. The vertex form provides a straightforward method for identifying properties of the parabola, such as its vertex.When dealing with quadratic equations, certain methods might be more advantageous, such as factoring, using the quadratic formula, or completing the square. The choice of method depends on the form of the quadratic given.In this specific exercise, solving for \(a\) involves substituting a known point into the vertex form equation. By setting this equal to the y-value of the point and isolating \(a\), we find the value needed to complete the equation:\[ -24 = a(0+6)^2 - 12 \] Solving the above for \(a\) shows how to find the missing coefficient, leading us to the final parabola equation.

Coordinate Geometry

Coordinate geometry is a fascinating branch of mathematics that utilizes coordinate planes to illustrate algebraic equations as graphs. Parabolas are just one of many curves studied in this field, represented through equations like the one we've dealt with.Here, the focus is on plotting the equation and understanding geometric properties of figures through their algebraic formulae.In our exercise, you can see how the vertex \((-6, -12)\) anchors the parabola, providing an easy starting point for graphing. This anchor point makes it simpler to visualize movements along the graph, aided by understanding how changes in the equation affect its shape. The point \((0, -24)\) further assists in verifying and refining our parabola equation within this coordinate plane. Coordinate geometry enables us to connect algebraic concepts with geometric visualization, providing a clearer, concrete understanding of mathematical relations.