Problem 33
Question
For the following exercises, use the vertex \((h, k)\) and a point on the graph \((x, y)\) to find the general form of the equation of the quadratic function. $$ (h, k)=(1,0),(x, y)=(0,1) $$
Step-by-Step Solution
Verified Answer
The general form is \(f(x) = x^2 - 2x + 1\).
1Step 1: Identify the Vertex Form
The vertex form of a quadratic function is given by:\[f(x) = a(x - h)^2 + k\]where \(h\) and \(k\) are the coordinates of the vertex of the parabola.
2Step 2: Substitute the Vertex Coordinates
Given that the vertex is \( (h, k) = (1, 0) \), substitute these values into the vertex form:\[f(x) = a(x - 1)^2 + 0\]which simplifies to:\[f(x) = a(x - 1)^2\]
3Step 3: Use the Point to Solve for 'a'
Use the given point \( (x, y) = (0, 1) \) to solve for \(a\). Substitute \(x = 0\) and \(y = 1\) into the equation:\[1 = a(0 - 1)^2\]\[1 = a(1)\]\[a = 1\]
4Step 4: Write the Quadratic Function in Vertex Form
Substitute \(a = 1\) back into the equation obtained in Step 2:\[f(x) = 1(x - 1)^2\]This is the vertex form of the quadratic function.
5Step 5: Convert to General Form
Expand the vertex form \(f(x) = 1(x - 1)^2\) to get the general form:\[f(x) = (x - 1)(x - 1)\]\[f(x) = x^2 - 2x + 1\]This is the general form of the quadratic function.
Key Concepts
Vertex FormParabolaGeneral Form
Vertex Form
A quadratic function can be expressed in various forms, and one of the most intuitive is the vertex form. The vertex form of a quadratic function is written as \(f(x) = a(x - h)^2 + k\). This form quickly reveals the vertex of the parabola, which is a crucial aspect of understanding the graph of a quadratic function. Here, the parameters \(h\) and \(k\) are the coordinates of the vertex, and they offer insights into the location of the graph's peak or valley.
Why is the vertex form useful? Here are a few reasons:
Why is the vertex form useful? Here are a few reasons:
- It is straightforward to identify the vertex of the parabola, which is the point \((h, k)\).
- The parameter \(a\) determines the direction and the width of the parabola. If \(a\) is positive, the parabola opens upward, and downward if negative.
- It effectively highlights transformations such as vertical shifts (by \(k\)), horizontal shifts (by \(h\)), and reflections based on the value of \(a\).
Parabola
A parabolic shape is the graph of any quadratic function. Parabolas have a characteristic 'U' or inverted 'U' shape and their appearance is controlled by the mathematical form of the equation representing them.
The most distinct features of a parabola include:
The turning point or vertex can determine real-life questions such as maximum height or minimum cost in optimization problems.
The most distinct features of a parabola include:
- The **vertex**, which is the highest or lowest point depending on whether it opens upward or downward, respectively.
- The **axis of symmetry**, a vertical line that divides the parabola into two mirror-image halves, passing through the vertex.
- The **focus** and **directrix**, which are geometric properties used to further describe the parabola's shape and position.
The turning point or vertex can determine real-life questions such as maximum height or minimum cost in optimization problems.
General Form
To fully analyze and graph a quadratic function, another form worth understanding is the general form, written as \(f(x) = ax^2 + bx + c\). While this form doesn't directly provide the vertex, it offers other insights:
The general form is particularly beneficial in calculus for deriving the quadratic equation, solving for roots, and facilitating the finding of critical points within various applications or when using differential concepts.
- The coefficient \(a\) influences the direction and steepness of the parabola—it determines whether the parabola opens up or down.
- The parameter \(b\) can affect the position of the vertex horizontally along the x-axis.
- \(c\) represents the y-intercept, or where the graph crosses the y-axis.
The general form is particularly beneficial in calculus for deriving the quadratic equation, solving for roots, and facilitating the finding of critical points within various applications or when using differential concepts.
Other exercises in this chapter
Problem 33
For the following exercises, use synthetic division to find the quotient. $$ \left(x^{4}-8 x^{3}+24 x^{2}-32 x+16\right) \div(x-2) $$
View solution Problem 33
For the following exercises, find the zeros and give the multiplicity of each. $$ f(x)=x^{2}\left(x^{2}+4 x+4\right) $$
View solution Problem 34
For the following exercises, use the given information to find the unknown value. \(y\) varies jointly as \(x\) and \(z\). When \(x=4\) and \(z=2\), then \(y=16
View solution Problem 34
For the following exercises, find the inverse of the function and graph both the function and its inverse. $$f(x)=(x-4)^{2}, x \geq 4$$
View solution