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 equation of the quadratic function is \( f(x) = x^2 - 2x + 1 \).
1Step 1: Write the Vertex Form of a Quadratic Function
The vertex form of a quadratic function is given by: \( f(x) = a(x-h)^2 + k \), where \((h, k)\) is the vertex of the parabola. Given \((h, k) = (1, 0)\), the vertex form becomes \( f(x) = a(x-1)^2 + 0 \) or simply \( f(x) = a(x-1)^2 \).
2Step 2: Substitute the Point into the Vertex Form
With the given point \((x, y) = (0, 1)\), substitute these values into the equation \( f(x) = a(x-1)^2 \). This gives \( 1 = a(0-1)^2 \).
3Step 3: Solve for the Coefficient a
From \( 1 = a(1) \), or \( 1 = a \), we solve for \( a \) and find that \( a = 1 \).
4Step 4: Write the Quadratic Function in General Form
Substitute \( a = 1 \) back into the vertex form equation to get \( f(x) = (x-1)^2 \). Expand this to the general form: \( f(x) = x^2 - 2x + 1 \).
Key Concepts
Vertex FormParabola VertexGeneral Form of a Quadratic Equation
Vertex Form
The vertex form of a quadratic function is a popular way to express a quadratic equation. This form is expressed as \( f(x) = a(x-h)^2 + k \). Here, \((h, k)\) represents the *vertex of the parabola*, which is the highest or lowest point of the graph depending on the direction it opens. The parameter \(a\) influences the width and the direction of the parabola. If \(a > 0\), the parabola opens upwards, whereas if \(a < 0\), it opens downwards.
The vertex form is particularly useful because it immediately gives the vertex \((h, k)\) of the parabola, making it easier to graph and analyze the function's behavior. For instance, when given a vertex \((1, 0)\), the vertex form becomes \( f(x) = a(x-1)^2 + 0 \). Therefore, it's easy to see where the parabola will be centered on the x-axis. This form can be transformed into other quadratic forms including the general form.
The vertex form is particularly useful because it immediately gives the vertex \((h, k)\) of the parabola, making it easier to graph and analyze the function's behavior. For instance, when given a vertex \((1, 0)\), the vertex form becomes \( f(x) = a(x-1)^2 + 0 \). Therefore, it's easy to see where the parabola will be centered on the x-axis. This form can be transformed into other quadratic forms including the general form.
Parabola Vertex
The parabola vertex is a key component of any quadratic function. It is the point \((h, k)\) that indicates the maximum or minimum value of the function. In terms of graphing, this is where the parabola changes direction.
With a quadratic function be expressed in vertex form \( f(x) = a(x-h)^2 + k \), the vertex \((h, k)\) is easy to identify. For instance, if the vertex is given as \((1, 0)\), the parabola reaches its peak or the lowest point at this location.
Understanding the vertex is crucial when sketching a parabola because it provides insight into the shape and position of the graph.
With a quadratic function be expressed in vertex form \( f(x) = a(x-h)^2 + k \), the vertex \((h, k)\) is easy to identify. For instance, if the vertex is given as \((1, 0)\), the parabola reaches its peak or the lowest point at this location.
- The x-coordinate \((h)\) represents the axis of symmetry of the parabola, which means the parabola is symmetric around the vertical line \(x = h\).
- The y-coordinate \((k)\) is the actual value at the vertex, giving a peak or a valley depending on the sign of \(a\).
Understanding the vertex is crucial when sketching a parabola because it provides insight into the shape and position of the graph.
General Form of a Quadratic Equation
The general form of a quadratic equation is given by \(f(x) = ax^2 + bx + c\). This format is perhaps the most familiar and useful for further algebraic operations like solving quadratic equations or factoring.
Starting from the vertex form, one can convert it into the general form by expanding the equation. For example, if you begin with the vertex form \(f(x) = (x-1)^2\), expanding this will yield a quadratic equation in general form: \(f(x) = x^2 - 2x + 1\).
The general form is versatile, allowing for easy computation of the x-intercepts using the quadratic formula if needed. It's a comprehensive way to understand the standard pattern of a parabola.
Starting from the vertex form, one can convert it into the general form by expanding the equation. For example, if you begin with the vertex form \(f(x) = (x-1)^2\), expanding this will yield a quadratic equation in general form: \(f(x) = x^2 - 2x + 1\).
- The coefficient \(a\) remains the same, controlling the direction and width of the parabola.
- The terms \(b\) and \(c\) adjust the position of the parabola on the graph.
The general form is versatile, allowing for easy computation of the x-intercepts using the quadratic formula if needed. It's a comprehensive way to understand the standard pattern of a parabola.
Other exercises in this chapter
Problem 33
For the following exercises, use synthetic division to find the quotient. Ensure the equation is in the form required by synthetic division. (Hint: divide the d
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. $$ f(x)=(x-4)^{2}, \quad x \geq 4 $$
View solution