Problem 55 - A Graphical Approach to Precalculus with Limits

Question

Find the equation of the quadratic function satisfying the given conditions. (Hint: Determine values of \(a\), \(h,\) and \(k\) that satisfy \(P(x)=a(x-h)^{2}+k .\) ) Express your answer in the form \(P(x)=a x^{2}+b x+c\). Use your calculator to support your results. Vertex \((-1,-4) ;\) through \((5,104)\)

Step-by-Step Solution

Answer

The equation is \(P(x) = 3x^{2} + 6x - 1\).

Step 1: Identify the Vertex

The vertex form of a quadratic function is given by \(P(x) = a(x-h)^{2} + k\). Here, the vertex is \((-1, -4)\), so we have \(h = -1\) and \(k = -4\). Substitute these values into the equation.

Step 2: Substitute Vertex in Vertex Form

Substitute \(h = -1\) and \(k = -4\) into the vertex form equation: \(P(x) = a(x + 1)^{2} - 4\).

Step 3: Substitute Known Point to Find 'a'

Use the point \((5, 104)\) to find the value of \(a\). Substitute \(x = 5\) and \(P(x) = 104\) into the equation: \[104 = a(5 + 1)^2 - 4\]Simplify and solve for \(a\):\[104 = a(36) - 4\]\[108 = 36a\]\[a = 3\]

Step 4: Substitute 'a' into Vertex Form

Now, substitute \(a = 3\) back into the vertex form equation:\[P(x) = 3(x + 1)^{2} - 4\]

Step 5: Expand to Standard Form

Expand the equation to get it in the form \(P(x) = ax^{2} + bx + c\):\[P(x) = 3(x^{2} + 2x + 1) - 4\]\[= 3x^{2} + 6x + 3 - 4\]\[= 3x^{2} + 6x - 1\]

Step 6: Verify the Solution

Verify by checking if the given vertex and point satisfy the equation. The vertex \((-1, -4)\) should be correct as in the vertex form, and substitute \((5, 104)\) back into \(P(x) = 3x^{2} + 6x - 1\) to check: \[3(5)^2 + 6(5) - 1 = 104\], which confirms the solution is correct.

Key Concepts

Understanding Vertex Form

The vertex form of a quadratic function provides a neat and efficient way to identify the key characteristics of a parabola. This form is written as \(P(x) = a(x-h)^2 + k\). Here, \(h\) and \(k\) represent the coordinates of the vertex. The vertex is a critical point that describes the peak or the lowest point of the parabola, depending on whether it opens upwards or downwards. The value of \(a\) affects the direction of the opening and the "width" of the parabola. A positive \(a\) makes the parabola open upwards, while a negative \(a\) makes it open downwards. The larger the absolute value of \(a\), the narrower the parabola.

Transitioning to Standard Form

The standard form of a quadratic equation is \(P(x) = ax^2 + bx + c\). This form allows for straightforward factoring and application of the quadratic formula. Starting from the vertex form, you can convert to the standard form by expanding the squared term and simplifying the equation.

For instance, given the vertex form \(P(x) = 3(x + 1)^2 - 4\), first expand \((x + 1)^2\) to get \(x^2 + 2x + 1\). Next, distribute the \(3\), resulting in:

• \(3(x^2 + 2x + 1) = 3x^2 + 6x + 3\)

Finally, subtract \(4\) to achieve the standard form:

• \(3x^2 + 6x + 3 - 4 = 3x^2 + 6x - 1\)

This simplified form is convenient for determining roots and analyzing symmetry.

Equation Solving Techniques

Solving quadratic equations involves finding the values of \(x\) that satisfy \(P(x) = 0\). Various methods include factoring, completing the square, or using the quadratic formula. For equations in standard form, the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) is a robust tool that works universally, regardless of whether the equation can be factored.

Let's say we have \(P(x) = 3x^2 + 6x - 1\). To solve using the quadratic formula, identify \(a = 3\), \(b = 6\), and \(c = -1\). Substitute these into the formula to find the solutions for \(x\). The discriminant \(b^2 - 4ac\) tells us about the nature of the roots:

• A positive discriminant implies two real and distinct roots.
• Zero indicates one real, repeated root.
• Negative means two complex roots.

Understanding these concepts enriches the experience of solving quadratic equations.

Finding the Vertex

The vertex of a parabola can be easily read from the vertex form, where it's explicitly written as \((h, k)\). If you're working with the standard form \(ax^2 + bx + c\), the vertex can be found using: \(h = \frac{-b}{2a}\). Once \(h\) is computed, substitute it back into the equation to find \(k\), confirming the vertex as \((h, k)\).

Suppose \(P(x) = 3x^2 + 6x - 1\), calculate \(h\):

• \(h = \frac{-6}{2*3} = -1\)

Then, substitute \(x = -1\) into \(P(x)\) to find \(k\):

• \(k = 3(-1)^2 + 6(-1) - 1 = -4\)

Therefore, the vertex is \((-1, -4)\). Knowing how to find and use these points is fundamental to graphing and understanding the shape and position of the parabola on the coordinate plane.