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How do you find x intercepts in a quadratic function?
The quadratic (parabola) intercepts the x-axis when y = 0. So substitute y=0 into y = f(x). Then you can solve for the x-values by any number of ways: Factoring, completing the square, or Quadratic Formula. It may turn out that the values of x which satisfies y=0 are complex {have an imaginary component}, which will tell you that the parabola does not have an x-intercept.
What are 3 other names for solutions of a quadratic equation?
Roots, zeroes, and x values are 3 other names for solutions of a quadratic equation.
Is x equals y a function?
y = x This is a line and a function. Function values are y values.
How many times will the graph of a quadratic function cross or touch the x-axis if the discriminant is zero?
It will touch it at exactly 1 point. If a quadratic function is given as f(x) = ax2 + bx + c, let the discriminant be denoted as D. Then the graph of y = f(x) will cross the x-axis at the x-values x = (-b + sqrt(D))/(2a) and x = (-b - sqrt(D))/(2a). When the discriminant D = 0, these 2 x-values are actually the same. Thus the graph will touch the x-axis only once.
How can you tell whether a graph is a function or not?
A vertical line test can be used to determine whether a graph is a function or not. If a vertical line intersects the graph more than once, then the graph is not a function.
What is 5x(squared)-16x-16?
It is a quadratic function of x. It takes different values which depend on the values given to x. It represents a parabola.
How can you use a graph to find zeros of a quadratic function?
The zeros of a quadratic function, if they exist, are the values of the variable at which the graph crosses the horizontal axis.
How do you plot the graph of a quadratic equation?
Just like any other equation, you can set up a table of x values, and calculate the corresponding y values. Then plot the points on the graph. In this case, it helps to have some familiarity with quadratic equations (you can find a discussion in algebra books), and recognize (from the form of the equation) whether your quadratic equation represents a parabola, a circle, an ellipse, or a hyperbola.
Is a quadratic function proportional?
I'd say no. A quadratic is not linear. It's slope is constantly changing, so there is not one proportion which can be used to predict future values.
Example of quadratic function?
x2+2x+1=y or y=x2 In this function the domain is x equals real values and the range is y equals all real values provided y is more than or equal to zero.
The graph of a a certain quadratic function has no x-intercepts. Which of the following are possible values for the discriminant check all that apply. (APEX)?
-7,-25
How do you make a table of values for a quadratic equation?
Simply learn and use the quadratic equation formula.
What is the solving for the roots of quadratic equation?
It is finding the values of the variable that make the quadratic equation true.
The graph of a certain quadratic function does not cross the x-axis Which of the following are possible values for the discriminant Check all that apply?
-1 -18 -25 -7
How do you get the points to graph a quadratic function?
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
How do you find x intercepts in a quadratic function?
The quadratic (parabola) intercepts the x-axis when y = 0. So substitute y=0 into y = f(x). Then you can solve for the x-values by any number of ways: Factoring, completing the square, or Quadratic Formula. It may turn out that the values of x which satisfies y=0 are complex {have an imaginary component}, which will tell you that the parabola does not have an x-intercept.
How do you create a function to calculate the roots of any given quadratic equation?
First rewrite the quadratic equation in the form: ax2 + bx + c = 0 where a , b and c are constant coefficients. Clearly, a is not = 0 for if it were then you would have a linear equation and not a quadratic. Then the roots of the quadratic are: x = [-b +/- sqrt(b2 - 4ac)]/2a where using the + and - values of the square root result in two solutions.
