Problem 78 - Algebra 2

Question

Each point lies on a parabola with vertex \((0,2) .\) Write the equation of the parabola. $$ (-2,8) $$

Step-by-Step Solution

Answer

The equation of the parabola is \( y=1.5x^2+2 \)

Step 1: Determine the vertex and the standard form

Given, the vertex of the parabola is (h,k) = (0,2). So, the equation of the parabola in vertex form is \( y=a(x-0)^2+2 = ax^2+2 \)

Step 2: Substituting the given point

The given point (-2,8) implies when x = -2 , y = 8. Substitute these values into the equation to solve for the value of 'a'. The equation then becomes: \( 8 = a(-2)^2 + 2 \)

Step 3: Solving for 'a'

Solving the above equation, we get: \( 8 = 4a + 2 \). Rearranging the equation and solving for 'a', we obtain: \( a = (8-2)/4 = 1.5 \)

Step 4: Writing the equation of parabola

Substitute 'a' back into the parabola equation. The equation of the parabola therefore is: \( y=1.5x^2+2 \)

Key Concepts

Vertex Form

The vertex form of a parabola is an important way to express the equation of a quadratic function. It's particularly useful because it easily tells us the location of the vertex, which is a key feature of the parabola. A parabola can open upwards or downwards and its shape can stretch or compress, but it will always have a singular vertex point (the highest or lowest point). The general vertex form of a quadratic equation is: \[ y = a(x - h)^2 + k \] Here,

• \((h, k)\) is the vertex of the parabola.
• "a" controls the width and the direction of the parabola. If \(a > 0\), it opens upwards; if \(a < 0\), it opens downwards.

In our example, the vertex is given as \((0, 2)\). Plugging these into the vertex form, we rewrite it as:\[ y = a(x - 0)^2 + 2 \] simplifying to:\[ y = ax^2 + 2 \]. This form sets the stage for finding the specific equation for this specific parabola by using additional information, like other points.

Substituting Coordinates

Once a parabola's vertex form is established, the next task is to find the value of "a" to specify the parabola completely. This is done by substituting a given point's coordinates into the vertex form equation. In this exercise, the given point is \((-2, 8)\). To substitute these coordinates:

• Replace \(x\) with -2 and \(y\) with 8 in the equation.

Thus, using the vertex form equation \( y = ax^2 + 2 \), we replace the values as follows:\[ 8 = a(-2)^2 + 2 \]Here, the formula becomes a simple arithmetic problem to solve for "a". This substitution transforms our problem into a straightforward calculation.

Solving Quadratic Equation

Having substituted the coordinates into the vertex form, we now solve for "a", completing the parabola's equation. Following the equation obtained from the substitution:\[ 8 = 4a + 2 \] Our goal is to isolate "a". Begin by subtracting 2 from both sides, resulting in:\[ 6 = 4a \]Next, divide both sides by 4 to solve for "a":\[ a = \frac{6}{4} = 1.5 \]With "a" found, the final step is to substitute back into the vertex form equation to find the specific equation for the parabola:\[ y = 1.5x^2 + 2 \]The parabola's equation is now fully determined and describes the specific curve that intersects with the vertex and passes through the given point. It shows that the parabola opens upwards and is stretched, as indicated by \(a = 1.5\).