You cannot, necessarily. Given a graph of the tan function, you could not.
A derivative of a function represents that equation's slope at any given point on its graph.
The Equation of a Rational Function has the Form,... f(x) = g(x)/h(x) where h(x) is not equal to zero. We will use a given Rational Function as an Example to graph showing the Vertical and Horizontal Asymptotes, and also the Hole in the Graph of that Function, if they exist. Let the Rational Function be,... f(x) = (x-2)/(x² - 5x + 6). f(x) = (x-2)/[(x-2)(x-3)]. Now if the Denominator (x-2)(x-3) = 0, then the Rational function will be Undefined, that is, the case of Division by Zero (0). So, in the Rational Function f(x) = (x-2)/[(x-2)(x-3)], we see that at x=2 or x=3, the Denominator is equal to Zero (0). But at x=3, we notice that the Numerator is equal to ( 1 ), that is, f(3) = 1/0, hence a Vertical Asymptote at x = 3. But at x=2, we have f(2) = 0/0, 'meaningless'. There is a Hole in the Graph at x = 2.
No, this is not a function. The graph would have a vertical line at x=-14. Since there are more than one y value for every given x value, the equation does not represent a function. The slope of the equation also does not exist.
Draw a graph of a given curve in the xoy plane. Now draw a vertical line so that it cuts the graph. If the vertical line cuts the graph in more than one ordinate then given graph is not a function. If it cuts the graph at a single ordinate such a graph is a function.(is called vertical line test)
A function must have a value for any given domain. For each edge (or interval), the sign graph has a sign (+ or -) . So, it is a function.
A derivative of a function represents that equation's slope at any given point on its graph.
A derivative of a function represents that equation's slope at any given point on its graph.
y = (some function of x). Since you've not given us the graph no-one can be any more helpful than that!
MATH 1003?
It can.
A linear function is a function whose graph is a straight line.
The Equation of a Rational Function has the Form,... f(x) = g(x)/h(x) where h(x) is not equal to zero. We will use a given Rational Function as an Example to graph showing the Vertical and Horizontal Asymptotes, and also the Hole in the Graph of that Function, if they exist. Let the Rational Function be,... f(x) = (x-2)/(x² - 5x + 6). f(x) = (x-2)/[(x-2)(x-3)]. Now if the Denominator (x-2)(x-3) = 0, then the Rational function will be Undefined, that is, the case of Division by Zero (0). So, in the Rational Function f(x) = (x-2)/[(x-2)(x-3)], we see that at x=2 or x=3, the Denominator is equal to Zero (0). But at x=3, we notice that the Numerator is equal to ( 1 ), that is, f(3) = 1/0, hence a Vertical Asymptote at x = 3. But at x=2, we have f(2) = 0/0, 'meaningless'. There is a Hole in the Graph at x = 2.
A function is an equation that is a straight line when plotted on a graph.
The answer will depend on the context. If the curve in question is a differentiable function then the gradient of the tangent is given by the derivative of the function. The gradient of the tangent at a given point can be evaluated by substituting the coordinate of the point and the equation of the tangent, though that point, is then given by the point-slope equation.
No, this is not a function. The graph would have a vertical line at x=-14. Since there are more than one y value for every given x value, the equation does not represent a function. The slope of the equation also does not exist.
If a graph is a function, it will always have y=... or x=... (or anoher letter equals an equation) for example y= 3x-12 is a function
If the function is a straight line equation that passes through the graph once, then that's a function, anything on a graph is a relation!