Problem 70 - Algebra for College Students

Question

a) Write the function \(y=3(x-2)^{2}+6\) in the form \(y=a x^{2}+b x+c,\) and find the vertex of the parabola using the formula \(x=\frac{-b}{2 a}\) b) Repeat part (a) with the functions \(y=-4(x-5)^{2}-9\) and \(y=3(x+2)^{2}-6\) c) What is the vertex for a parabola that is written in the form \(y=a(x-h)^{2}+k ?\) Explain your answer.

Step-by-Step Solution

Answer

For \(y = 3(x-2)^2 + 6\), the vertex is \((2, 6)\). For \(y = -4(x-5)^2 - 9\), the vertex is \((5, -9)\). For \(y = 3(x+2)^2 - 6\), the vertex is \((-2, -6)\). Vertex form \(y = a(x-h)^2 + k\) has a vertex at \((h, k)\).

Step 1: Expand the given function

For the function \(y = 3(x-2)^2 + 6\), expand the expression \((x-2)^2\) and then distribute the 3.

Step 2: Distribute and combine like terms

\((x-2)^2 = x^2 - 4x + 4\). So, \(y = 3(x^2 - 4x + 4) + 6\). Now distribute 3 and combine terms: \(y = 3x^2 - 12x + 12 + 6\). Thus, \(y = 3x^2 - 12x + 18\).

Step 3: Find the vertex using \(x=-\frac{b}{2a}\)

The vertex form of a parabola can be found using the formula \(x = -\frac{b}{2a}\). Here, \(a = 3\) and \(b = -12\). Thus, \(x = -\frac{-12}{2(3)} = 2\). Substitute \(x = 2\) back in the equation \(y = 3(2)^2 - 12(2) + 18 = 6\). Vertex is at \((2, 6)\).

Step 4: Repeat for \(y = -4(x-5)^2 - 9\)

Expand \(y = -4(x-5)^2 - 9\) to standard form: \((x-5)^2 = x^2 - 10x + 25\). Thus, \(y = -4(x^2 - 10x + 25) - 9 = -4x^2 + 40x - 100 - 9 = -4x^2 + 40x - 109\). The vertex can be found using \(x = -\frac{b}{2a} = -\frac{40}{-8} = 5\). Substituting \(x = 5\) back: \(y = -4(5)^2 + 40(5) - 109 = -9\). Vertex is at \((5, -9)\).

Step 5: Repeat for \(y = 3(x+2)^2 - 6\)

Expand \((x+2)^2 = x^2 + 4x + 4\). Thus, \(y = 3(x^2 + 4x + 4) - 6 = 3x^2 + 12x + 12 - 6 = 3x^2 + 12x + 6\). Vertex form: \(x = -\frac{b}{2a} = -\frac{12}{6} = -2\). Substituting \(x = -2\) back: \(y = 3(-2)^2 + 12(-2) + 6 = -6\). Vertex is at \((-2, -6)\).

Step 6: Finding vertex from \(y = a(x-h)^2 + k\)

The vertex of a parabola is at \((h, k)\) when the function is in the form \(y = a(x-h)^2 + k\). This is because when \(x = h\), the squared term becomes zero, leaving \(y = k\). Hence, the vertex is \((h, k)\) in this form.

Key Concepts

Expanding Quadratic Functions

When you expand a quadratic function, you transform it from a more compressed vertex form into a wider, expanded standard form.
Let's illustrate this using an example. Consider the function:
\(y = 3(x-2)^2 + 6\).
First, you need to expand the squared term, \( (x-2)^2\). When you do this, you get \(x^2 - 4x + 4\).
Now, substitute this back into the equation:

\(y = 3(x^2 - 4x + 4) + 6\).

Next, distribute the 3 across the terms inside the parentheses:
\(y = 3x^2 - 12x + 12 + 6\).

Finally, combine the constants:
\(y = 3x^2 - 12x + 18\).
Thus, we've expanded \(y = 3(x-2)^2 + 6\) into its standard form, \(y = 3x^2 - 12x + 18\).
This process helps us better understand the shape and orientation of the parabola in standard form.

Standard Form of a Quadratic Equation

The standard form of a quadratic equation is written as:
\(y = ax^2 + bx + c\).
This format helps to easily identify the coefficients that influence the parabola's shape and position.
Here's how you find these coefficients from the equations provided in the exercise.
For the function \(y = -4(x-5)^2 - 9,\) you expand and simplify it to get:
\(y = -4x^2 + 40x - 109\). Converting the initial equation into standard form involves:
Expanding the squared term \( (x-5)^2 \), distributing the constant,
and combining like terms.

• a is the coefficient of \(x^2 \),
• b is the coefficient of \(x \), and
• c is the constant term.

These coefficients are crucial. 'a' affects whether the parabola opens upwards or downwards; 'b' influences the axis of symmetry; and 'c' is the y-intercept. Understanding these coefficients helps in analyzing the quadratic function's behavior.

Vertex Form Equation

The vertex form of a quadratic equation is:
\(y = a(x-h)^2 + k\).
This form is particularly useful for easily identifying the vertex of the parabola.
The vertex \((h, k)\) is the point where the parabola changes direction. Let’s break it down with an example.
For the function \(y = 3(x+2)^2 - 6\),
it is already in vertex form.
From this, we can tell the vertex immediately:
\((h, k) = (-2, -6)\). The vertex form makes it straightforward to pick out the vertex,
because when \(x = h\), the squared part \(x - h\) becomes zero.
Hence, the value of \(y = k\). Moving from standard form back to vertex form requires completing the square,
but starting with the vertex form can make many problems simpler.
This approach can also help in graphing parabolas and understanding their transformations better.