Problem 39
Question
The graph of each equation is to be translated 3 units right and 5 units up. Write each new equation. \(y=4 x^{2}\)
Step-by-Step Solution
Verified Answer
The translated equation is \(y = 4(x - 3)^{2} + 5\).
1Step 1: Translate the graph to the right
In a graphical interpretation, moving the graph 3 units to the right means that the value of x in the equation will decrease by 3. To represent this mathematically, replace every instance of x in the equation with (x - 3). This changes the equation \(y = 4x^{2}\) to \(y = 4(x - 3)^{2}\).
2Step 2: Translate the graph upwards
Moving the graph 5 units up means that the value of y in the equation will increase by 5. To illustrate this numerically, add 5 to the entire equation, changing it from \(y = 4(x - 3)^{2}\) to \(y = 4(x - 3)^{2} + 5\). This completes the translation of the graph.
Key Concepts
Translating FunctionsQuadratic FunctionsGraphing Equations
Translating Functions
Translating functions involves shifting the entire graph of a function either horizontally or vertically, without altering its shape or orientation.
To translate a function:
To translate a function:
- Horizontally: Adjust the x-values in the equation. Moving the function to the right involves subtracting a value from x, while moving it to the left involves adding to x.
- Vertically: Change the y-values by adding or subtracting a constant. To move a function up, a positive constant is added, whereas shifting it down requires subtracting a constant.
Quadratic Functions
A quadratic function is a type of polynomial function where the highest degree of the variable is 2. It follows the form \[ f(x) = ax^2 + bx + c \]where:
Depending on the coefficient a, the parabola can open either upward (when a > 0) or downward (when a < 0). The backbone of our example is the function \(y = 4x^2\), a simple form of a quadratic equation with a vertex at the origin (0, 0) and opens upwards because the coefficient 4 is positive.
When we translate this function, the shape remains a parabola, but its position on the coordinate plane changes. Applying the translations as described, the function becomes \(y = 4(x - 3)^2 + 5\), shifting both right and up from its original placement.
- a, b, and c are constants;
- a \( eq \) 0;
- x is the variable.
Depending on the coefficient a, the parabola can open either upward (when a > 0) or downward (when a < 0). The backbone of our example is the function \(y = 4x^2\), a simple form of a quadratic equation with a vertex at the origin (0, 0) and opens upwards because the coefficient 4 is positive.
When we translate this function, the shape remains a parabola, but its position on the coordinate plane changes. Applying the translations as described, the function becomes \(y = 4(x - 3)^2 + 5\), shifting both right and up from its original placement.
Graphing Equations
Graphing equations is a visual way of representing a mathematical function or relationship.
For quadratic functions, graphing primarily involves sketching the shape of a parabola. Here’s how you can graph a quadratic function effectively:
For quadratic functions, graphing primarily involves sketching the shape of a parabola. Here’s how you can graph a quadratic function effectively:
- Identify the vertex, which is the turning point of the parabola. For functions like \(y = ax^2\), the vertex is at the origin (0,0). For functions like our translated example \(y = 4(x - 3)^2 + 5\), the vertex becomes (3,5).
- Determine if the parabola opens upward or downward based on the sign of coefficient a. A positive a means it opens upwards.
- Select a few key points around the vertex to plot for accuracy. Substitute x-values into the equation to find corresponding y-values.
- Draw the axis of symmetry, a vertical line that goes through the vertex. This helps ensure that your parabola is symmetric.
- Connect the plotted points smoothly to form the parabola shape.
Other exercises in this chapter
Problem 38
Write an equation of a parabola with a vertex at the origin. \(\operatorname{directrix} y=5\)
View solution Problem 38
Write the equation of the circle that passes through the given point and has a center at the origin. (Hint: You can use the distance formula to find the radius.
View solution Problem 39
Writing. Describe the similarities and differences between hyperbolas and ellipses.
View solution Problem 39
Write an equation of a parabola with a vertex at the origin. focus at \((-7,0)\)
View solution