Problem 1
Question
For Problems 1 through 8, graph the function. Label the \(x\) - and \(y\) -intercepts and the coordinates of the vertex. $$ f(x)=-x^{2}+1 $$
Step-by-Step Solution
Verified Answer
The \(y\)-intercept is at (0,1), the \(x\)-intercepts are at (-1,0) and (1,0), and the vertex is at (0,1). Based on these points and the shape of a quadratic function, a graph of the function can be drawn.
1Step 1: Find the \(y\)-intercept
The \(y\)-intercept of a function is where the curve crosses the \(y\)-axis. In other words, it is the value of the function when \(x=0\). If we substitute \(x=0\) into our function \(f(x)=-x^{2}+1\), we get \(f(0) = -(0)^{2}+1 = 1\). Thus, the \(y\)-intercept is at (0,1)
2Step 2: Find the \(x\)-intercepts
The \(x\)-intercepts of a function is where the curve crosses the \(x\)-axis. These points satisfy the equation \(f(x)=0\). So, we set the function equal to zero and solve for \(x\):\[0=-x^{2}+1 ⇒ -x^{2} = -1 ⇒ x^{2} = 1 ⇒ x = ±1\]Thus, the \(x\)-intercepts are at (-1,0) and (1,0)
3Step 3: Find the Vertex
The vertex of a parabola \(y=ax^{2}+bx+c\) is given by \((-b/2a , f(-b/2a))\). But in \(-x^{2}+1\), the coefficient of \(x\) is 0 (i.e., \(b=0\)), and the coefficient of \(x^{2}\) is -1 (i.e., \(a=-1\)). Therefore, the vertex is located at \((0, f(0)) = (0,1)\)
4Step 4: Graphing the Function
Now, taking all above information together, plot the points and sketch a smooth curve passing through them, making sure it open downwards because the coefficient of \(x^2\) is negative, which represents a downward facing parabola. The vertex is the maximum point on the curve located at (0,1), and the \(x\)-intercepts are (-1,0) and (1,0). Plot this on a graph
Key Concepts
Understanding x-intercepts in Graphing FunctionsExploring y-intercepts in ParabolasDiscovering the Vertex of a Parabola
Understanding x-intercepts in Graphing Functions
The x-intercepts of a function are the points where the graph crosses the x-axis. These points occur when the output of the function is zero, meaning we set the function equal to zero to find them. For a quadratic function like \( f(x) = -x^2 + 1 \), we solve the equation \( f(x) = 0 \).
This becomes \( -x^2 + 1 = 0 \), which rearranges to \( x^2 = 1 \). Solving \( x^2 = 1 \) gives \( x = \pm 1 \).
Thus, the x-intercepts are located at the points (-1, 0) and (1, 0). These intercepts provide valuable information about where the parabola touches the x-axis and are crucial for drawing the graph accurately.
This becomes \( -x^2 + 1 = 0 \), which rearranges to \( x^2 = 1 \). Solving \( x^2 = 1 \) gives \( x = \pm 1 \).
Thus, the x-intercepts are located at the points (-1, 0) and (1, 0). These intercepts provide valuable information about where the parabola touches the x-axis and are crucial for drawing the graph accurately.
Exploring y-intercepts in Parabolas
The y-intercept is where the graph crosses the y-axis. To find the y-intercept of the function \( f(x) = -x^2 + 1 \), substitute \( x = 0 \) into the equation.
Doing so gives \( f(0) = -(0)^2 + 1 = 1 \). This means the y-intercept is at the point (0, 1).
The y-intercept gives us a starting point when graphing a function. It shows how the graph behaves in respect to the y-axis and is essential for understanding the general position of the parabola in the coordinate plane.
Doing so gives \( f(0) = -(0)^2 + 1 = 1 \). This means the y-intercept is at the point (0, 1).
The y-intercept gives us a starting point when graphing a function. It shows how the graph behaves in respect to the y-axis and is essential for understanding the general position of the parabola in the coordinate plane.
Discovering the Vertex of a Parabola
In a quadratic function, the vertex is a key feature that represents the highest or lowest point of the parabola. For the function \( f(x) = -x^2 + 1 \), since there is no linear term (\(b=0\)), finding the vertex is straightforward.
The standard form of a quadratic function is \( ax^2 + bx + c \). The formula to find the vertex is \( (-b/2a, f(-b/2a)) \). Here, \( a = -1 \) and \( b = 0 \), leading to the vertex \((0, f(0))\). Calculating \( f(0) \) gives us \( 1 \), so the vertex is at (0, 1).
This point is also where the y-intercept is, which signifies the peak of the parabola opening downwards due to a negative \(a\) value. Understanding the vertex helps in sketching the curve and knowing its maximum or minimum point.
The standard form of a quadratic function is \( ax^2 + bx + c \). The formula to find the vertex is \( (-b/2a, f(-b/2a)) \). Here, \( a = -1 \) and \( b = 0 \), leading to the vertex \((0, f(0))\). Calculating \( f(0) \) gives us \( 1 \), so the vertex is at (0, 1).
This point is also where the y-intercept is, which signifies the peak of the parabola opening downwards due to a negative \(a\) value. Understanding the vertex helps in sketching the curve and knowing its maximum or minimum point.
- Vertex: Identifies the turning point of the parabola.
- Location: At coordinates (0, 1) in this example.
- Significance: Max point here due to downward opening.
Other exercises in this chapter
Problem 1
Let \(f(x)=x+3, g(x)=x-5, h(x)=f(x) g(x)\), and \(j(x)=\frac{f(x)}{g(x)}\). Solve the following equations. Find all \(x\) that satisfy the equation. (a) \(h(x)=
View solution Problem 1
A seat on a round-trip charter flight to Cairo costs \(\$ 720\) plus a surcharge of \(\$ 10\) for every unsold seat on the airplane. (If there are 10 seats left
View solution Problem 2
For Problems 1 through 8, graph the function. Label the \(x\) - and \(y\) -intercepts and the coordinates of the vertex. $$ f(x)=(x+3)(2 x-6) $$
View solution