The question implies that it is a choice between two or more examples. In those circumstances would it be too much to expect that you make sure that there are some graphs to select from?
the graph is moved down 6 units
y=x2+4x+1
x2+(y-x2/3)2=1
y=x2
1
9
Therefore x2=9+y2. And x is the square-root of that (with two values plus and minus). Choose a value of y, and work out x2 and therefore the values of x. Plot the two (+ and -) on a graph and continue for more values of y.
No translation will invert a quadratic graph.
First, reflect the graph of y = x² in the x-axis (line y = 0) to obtain the graph of y = -x²; then second, shift it 3 units up to obtain the graph of y = -x² + 3.
7
the graph is moved down 6 units
y=x2+4x+1
x2+(y-x2/3)2=1
y=x2
1
y = - x2 +6x - 5.5
y = x2 + 6x + 7a = 1, b = 6, c = 7Since a is positive, the graph opens upward. You can find the vertex coordinates (-b/2a, f(-b/2a)) = (-3, f(-3)) = (-3, -2), so the equation of the axis of symmetry is x = -b/2a = -3, plot the y-intercept point (0, 7), plot the point (-b/a, 7) = (-6, 7), and draw the graph that passes through these points.Or complete the square.y = x2 + 6x + 7y = x2 + 6x + 9 - 9 + 7y = (x2 + 6x + 9) - 2y = (x + 3)2 - 2So start with the graph of y = x2 whose vertex is at the origin. Move it 3 units to the left, and 2 units do