point
If the graph is a function, no line perpendicular to the X-axis can intersect the graph at more than one point.
x=0
It is impossible to determine the function here. (-1,4) is just a point on a graph.
If for every point on the horizontal axis, the graph has one and only one point corresponding to the vertical axis; then it represents a function. Functions can not have discontinuities along the horizontal axis. Functions must return unambiguous deterministic results.
The answer depends on the form in which the quadratic function is given. If it is y = ax2 + bx + c then the x-coordinate of the turning point is -b/(2a)
vertex
When the graph of a quadratic crosses the x-axis twice it means that the quadratic has two real roots. If the graph touches the x-axis at one point the quadratic has 1 repeated root. If the graph does not touch nor cross the x-axis, then the quadratic has no real roots, but it does have 2 complex roots.
No vertical line will intersect the graph in more than one point. The fundamental flaw is that no graph can show that it does not happen beyond the domain of the graph.
Depending on the graph, for a quadratic function the salient features are: X- intercept, Y-intercept and the turning point.
...i need the answer to that too...
If the graph of the function is a continuous line then the function is differentiable. Also if the graph suddenly make a deviation at any point then the function is not differentiable at that point . The slope of a tangent at any point of the graph gives the derivative of the function at that point.
The quadratic equation is y=ax^2 +bx +c. So, you substitute in the values of a, b, and c to the quadratic formula (x= -b +/- \|b^2-4ac all over 2a) in order to find the x value then, substitute in x to the quadratic equation and solve. You will have point (x,y) to graph
A zero of a function is a point at which the value of the function is zero. If you graph the function, it is a point at which the graph touches the x-axis.
Apex.
zero
If the point (x,y) is on the graph of the even function y = f(x) then so is (-x,y)