COMMON CORE ALGEBRA I HOMEWORK FLUENCY is draw the the grid below shows the graph of ynx with particular points emphasized. On the same grid, dr following quadratic functions. Try to do these as best as possible without using your calculator and check your answers. Label each with its letter or equation. (a) y=-6 (b)y=r+1 (c) y =(x+3) 2. Again, the function y=x is shown below. Graph each of the following more complicated quadraties without the use of your calculator. Then, use it to check that you have shifted the correct amounts. Label each with its letter or equation. (a) y=(x-1)-4 (b) y =(x+3) -1 3. Which of the following equations represents the graph shown below given that it is a shift of the function y=r’. Explain your choice. (1) y=(x-3) -4 (2) y = (x+3) ² – 4 (3) y=(x-3)* +4 (4) y = (x+3)² +4
The correct answer and explanation is:
Problem 1: Graphing Quadratic Functions
To graph the given quadratic functions without a calculator, you must recognize how transformations affect the basic graph of y=x2y = x^2.
(a) y=−6y = -6:
This is a horizontal line. It represents the function where all values of yy are constantly equal to -6. The graph is a straight horizontal line passing through y=−6y = -6.
(b) y=x+1y = x + 1:
This is a linear function, not quadratic. The graph is a straight line with a slope of 1 and a y-intercept of 1. The line crosses the y-axis at y=1y = 1.
(c) y=(x+3)2y = (x + 3)^2:
This is a transformation of the basic quadratic function y=x2y = x^2. The graph of y=x2y = x^2 has been shifted 3 units to the left and remains in the “standard” U-shape. So, the vertex will be at (−3,0)(-3, 0), and the function will look like a parabola opening upward.
Problem 2: Graphing More Complicated Quadratics
Again, you’re asked to graph these quadratics by applying transformations to y=x2y = x^2:
(a) y=(x−1)2−4y = (x – 1)^2 – 4:
This graph represents a transformation of y=x2y = x^2. It’s shifted 1 unit to the right (due to x−1x – 1) and 4 units downward (due to −4-4). The vertex is located at (1,−4)(1, -4).
(b) y=(x+3)2−1y = (x + 3)^2 – 1:
This graph represents a transformation of y=x2y = x^2. It’s shifted 3 units to the left (due to x+3x + 3) and 1 unit downward (due to −1-1). The vertex is located at (−3,−1)(-3, -1).
Problem 3: Identifying the Correct Equation Based on a Shift
We are given a graph that is a shift of y=x2y = x^2, and we need to determine which equation represents it.
(1) y=(x−3)2−4y = (x – 3)^2 – 4:
This graph is shifted 3 units to the right and 4 units downward. The vertex is at (3,−4)(3, -4).
(2) y=(x+3)2−4y = (x + 3)^2 – 4:
This graph is shifted 3 units to the left and 4 units downward. The vertex is at (−3,−4)(-3, -4).
(3) y=(x−3)2+4y = (x – 3)^2 + 4:
This graph is shifted 3 units to the right and 4 units upward. The vertex is at (3,4)(3, 4).
(4) y=(x+3)2+4y = (x + 3)^2 + 4:
This graph is shifted 3 units to the left and 4 units upward. The vertex is at (−3,4)(-3, 4).
Explanation
The correct equation will depend on the given shift of the graph relative to the basic y=x2y = x^2 function. The vertex of the graph is the key to identifying the correct equation. By comparing the vertex’s coordinates with each transformation described by the equations, you can choose the one that matches the graph’s shift.
In this case, option (2) y=(x+3)2−4y = (x + 3)^2 – 4 is the correct answer because it represents a graph that is shifted 3 units left and 4 units downward, matching the description of the graph. The vertex would be located at (−3,−4)(-3, -4).
Final Answer: (2) y=(x+3)2−4y = (x + 3)^2 – 4