Тема: Help on my math homework (quadratic formula)?

A quadratic function can be graphed using a table of values. The graph creates a parabola. The parabola contains specific points, the vertex, and up to two zeros or x-intercepts. The zeros are the points where the parabola crosses the x-axis.

Given the following function:
h(x) = 1/2x 2 - x +2

  • Predict whether the parabola will open up or down.
  • Find the x coordinate of the vertex using the vertex formula.
  • Create a table of values and graph the parabola.
  • Identify the zeros of function.

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2

x = [5±sqrt(73)]/4

It may help to start by graphing the function on your calculator (assuming you are allowed to use a calculator) by entering the function as it appears into y1. Set the window values to center around the origin. Example: both x and y values range from -25 to 25. A graph of the function shows that it does not cross the x-axis; there are no real zeros and no x-intercept. You can show this mathematically by expanding the function. f(x)= -2 (x-4)(x-4) - 18 f(x)= -2 (x^2-4x-4x+16) - 18 f(x)= -2x^2+16x-32-18 f(x)=-2x^2+16x-50 Because the leading coefficient of the quadratic (-2) is negative, the function is concave down. It has a vertical translation that shifted it down 50 units as well, meaning that the vertex (peak) of the function occurs at x=-50, far below the x-axis. You can see these values in your graph of the function. To solve for y-intercepts, simply set the function equal to zero. f(x)=-2(x-4)-18=0 Solve this equality for x; the value of x indicates the point on the graph where the function crosses the y-axis. Unfortunately, as far as I am aware, there are no y-intercepts for your function either. This is entirely possible, but also may just be because I misread your f(x) equation. I did the math by hand & also checked it with a special program in my calculator; both indicated that there are no real solutions.

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A quadratic function can be graphed using a table of values. The graph creates a parabola. The parabola contains specific points, the vertex, and up to two zeros or x-intercepts. The zeros are the points where the parabola crosses the x-axis.

If the coefficient of the squared term is positive, the parabola opens up. The vertex of this parabola is called the minimum point.

I assume that the equation is 3x^2 + 12x + 7 = 0 The Quadratic Formula: for ax^2 + bx + c = 0: x = (-b + or - SQR(b^2 - 4ac))/2a so x = (-12 + or - SQR(144 - 4(3)(7)))/2(3) x = (-12 + or - SQR(144 - 84))/6 x = (-12 + or - SQR(60))/6 simplify as desired, you will get two answers called roots. The second part of your question makes no sense, there is no y in the equation.

Home » Functions » Vertex Formula Using the Vertex Formula Quadratic Functions - Lesson 2. Before we begin this lesson on using the vertex formula, let's briefly.

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