the graph is moved down 6 units
There are none. For this equation, there is nonreal answer, as the graph of the quadratic does not pass below the x-axis
There are no real solutions to this equation because you cannot take the square root of a negative number. However,x2 + 4 = 0x2 = -4sqrt(x2) = sqrt(-4)x = 2i, -2ihere are the imaginary solutions.
It will have two solutions because its a quadratic equation: x = -8.472135955 or x = 0.472135955
No, because there is more than one solution: y2 = x2 y = ±(x2)1/2 y = ±x Because there are multiple solutions for a single value of x, this does not qualify as a function.
No translation will invert a quadratic graph.
the graph is moved down 6 units
Let's say you have the quadratic equation x2 - 7x + 12 = 0. Plot the graph of y = x2 - 7x + 12. Where y = 0 (when the graph crosses the x-axis) is a solution to the equation. In this case, it crosses at the points (3,0) & (4,0) so the solutions are x = 3 and x = 4. Now if the graph never touches the x-axis, that means the solutions to the equation are complex numbers.
There are none. For this equation, there is nonreal answer, as the graph of the quadratic does not pass below the x-axis
There are no real solutions to this equation because you cannot take the square root of a negative number. However,x2 + 4 = 0x2 = -4sqrt(x2) = sqrt(-4)x = 2i, -2ihere are the imaginary solutions.
A graph that has 1 parabolla that has a minimum and 1 positive line.
It will have two solutions because its a quadratic equation: x = -8.472135955 or x = 0.472135955
One way would be to graph the two equations: the parabola y = x² + 4x + 3, and the straight line y = 2x + 6. The two points where the straight line intersects the parabola are the solutions. The 2 solution points are (1,8) and (-3,0)
No, because there is more than one solution: y2 = x2 y = ±(x2)1/2 y = ±x Because there are multiple solutions for a single value of x, this does not qualify as a function.
9
x2≤64
The graph is a parabola facing (opening) upwards with the vertex at the origin.