x-2 = 3 is the single point x = 5. Not quite sure of the point in graphing a single value of a single variable!
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.
x2≤64
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.
A graph that has 1 parabolla that has a minimum and 1 positive line.
There are none. For this equation, there is nonreal answer, as the graph of the quadratic does not pass below the x-axis
9
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)
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.
x2≤64
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.
It will have two solutions because its a quadratic equation: x = -8.472135955 or x = 0.472135955
The graph is a parabola facing (opening) upwards with the vertex at the origin.