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The graph of a quadratic equation of the form ax + bx + c or a (x - h) + k is a parabola (U-shaped curve). To plot such an equation, you need to find the vertex of the parabola, its direction and points of intersection with the X and Y axes. If you are given a relatively simple quadratic equation, then you can substitute different values ​​of "x" into it, find the corresponding values ​​of "y" and build a graph ...

Steps

1. 1 The quadratic equation can be written in a standard form and in a non-standard form. You can use any kind of equation to plot a quadratic equation (the plotting method is slightly different). As a rule, in problems, quadratic equations are given in a standard form, but this article will tell you about both types of writing a quadratic equation.
   • Standard form: f (x) = ax + bx + c, where a, b, c are real numbers and a ≠ 0.
     • For example, two equations of the standard form: f (x) = x + 2x + 1 and f (x) = 9x + 10x -8.
   • Non-standard form: f (x) = a (x - h) + k, where a, h, k are real numbers and a ≠ 0.
     • For example, two equations of a non-standard form: f (x) = 9 (x - 4) + 18 and -3 (x - 5) + 1.
   • To plot a quadratic equation of any kind, you first need to find the vertex of the parabola, which has coordinates (h, k). The coordinates of the vertex of the parabola in the equations of the standard form are calculated by the formulas: h = -b / 2a and k = f (h); the coordinates of the vertex of the parabola in equations of a non-standard form can be obtained directly from the equations.
2. 2 To plot a graph, you need to find the numerical values ​​of the coefficients a, b, c (or a, h, k). In most problems, quadratic equations are given with numerical values ​​of the coefficients.
   • For example, in the standard equation f (x) = 2x + 16x + 39 a = 2, b = 16, c = 39.
   • For example, in a non-standard equation f (x) = 4 (x - 5) + 12, a = 4, h = 5, k = 12.
3. 3 Calculate h in the standard equation (in the non-standard it is already given) using the formula: h = -b / 2a.
   • In our standard equation example, f (x) = 2x + 16x + 39 h = -b / 2a = -16/2 (2) = -4.
   • In our example of a non-standard equation, f (x) = 4 (x - 5) + 12 h = 5.
4. 4 Calculate k in the standard equation (in the non-standard it is already given). Remember that k = f (h), that is, you can find k by substituting the found value for h instead of "x" in the original equation.
   • You found that h = -4 (for the standard equation). To calculate k, substitute this value for "x":
     • k = 2 (-4) + 16 (-4) + 39.
     • k = 2 (16) - 64 + 39.
     • k = 32 - 64 + 39 = 7
   • In a non-standard equation, k = 12.
5. 5 Draw a vertex with coordinates (h, k) on the coordinate plane. h is plotted along the X-axis and k is plotted along the Y-axis. The top of a parabola is either the lowest point (if the parabola is pointing up) or the highest point (if the parabola is pointing down).
   • In our standard equation example, the vertex has coordinates (-4, 7). Draw this point on the coordinate plane.
   • In our example of a custom equation, the vertex has coordinates (5, 12). Draw this point on the coordinate plane.
6. 6 Draw the axis of symmetry of the parabola (optional). The axis of symmetry passes through the apex of the parabola parallel to the Y axis (that is, strictly vertical). The axis of symmetry divides the parabola in half (that is, the parabola is mirror-symmetrical about this axis).
   • In our standard equation example, the axis of symmetry is a straight line parallel to the Y axis and passing through the point (-4, 7). Although this line is not part of the parabola itself, it gives an idea of ​​the symmetry of the parabola.
7. 7 Determine the direction of the parabola - up or down. This is very easy to do.If the coefficient "a" is positive, then the parabola is directed upward, and if the coefficient "a" is negative, then the parabola is directed downward.
   • In our example of the standard equation f (x) = 2x + 16x + 39, the parabola is pointing up, since a = 2 (positive coefficient).
   • In our example of a non-standard equation f (x) = 4 (x - 5) + 12, the parabola is also directed upwards, since a = 4 (positive coefficient).
8. 8 If necessary, locate and plot the x-intercept. These points will help you a lot when drawing a parabola. There can be two, one or none (if the parabola is directed upwards and its vertex lies above the X-axis, or if the parabola is directed downwards and its vertex lies below the X-axis). To calculate the coordinates of the points of intersection with the X-axis, do the following:
   • Set the equation to zero: f (x) = 0 and solve it. This method works with simple quadratic equations (especially non-standard ones), but can be extremely difficult for complex equations. In our example:
     • f (x) = 4 (x - 12) - 4
     • 0 = 4 (x - 12) - 4
     • 4 = 4 (x - 12)
     • 1 = (x - 12)
     • √1 = (x - 12)
     • +/- 1 = x -12. The points of intersection of the parabola with the X-axis have coordinates (11,0) and (13,0).
   • Factor the standard-form quadratic equation: ax + bx + c = (dx + e) ​​(fx + g), where dx × fx = ax, (dx × g + fx × e) = bx, e × g = c. Then set each binomial to 0 and find the values ​​for "x". For example:
     • x + 2x + 1
     • = (x + 1) (x + 1)
     • In this case, there is a single point of intersection of the parabola with the x-axis with coordinates (-1,0), because at x + 1 = 0 x = -1.
   • If you cannot factor the equation, solve it using the quadratic formula: x = (-b +/- √ (b - 4ac)) / 2a.
     • For example: -5x + 1x + 10.
     • x = (-1 +/- √ (1 - 4 (-5) (10))) / 2 (-5)
     • x = (-1 +/- √ (1 + 200)) / - 10
     • x = (-1 +/- √ (201)) / - 10
     • x = (-1 +/- 14.18) / - 10
     • x = (13.18 / -10) and (-15.18 / -10). The points of intersection of the parabola with the X axis have coordinates (-1,318,0) and (1,518,0).
     • In our example, the equations of the standard form 2x + 16x + 39:
     • x = (-16 +/- √ (16 - 4 (2) (39))) / 2 (2)
     • x = (-16 +/- √ (256 - 312)) / 4
     • x = (-16 +/- √ (-56) / - 10
     • Since it is impossible to extract the square root of a negative number, in this case the parabola does not intersect the X-axis.
9. 9 Locate and plot the y-intercept as needed. This is very easy - plug x = 0 into the original equation and find the value "y". The Y-intercept is always the same. Note: in equations of the standard form, the intersection point has coordinates (0, s).
   • For example, the parabola of the quadratic equation 2x + 16x + 39 intersects with the Y-axis at the point with coordinates (0, 39), since c = 39. But this can be calculated:
     • f (x) = 2x + 16x + 39
     • f (x) = 2 (0) + 16 (0) + 39
     • f (x) = 39, that is, the parabola of this quadratic equation intersects the Y-axis at the point with coordinates (0, 39).
   • In our example of a non-standard equation 4 (x - 5) + 12, the y-intercept is calculated as follows:
     • f (x) = 4 (x - 5) + 12
     • f (x) = 4 (0 - 5) + 12
     • f (x) = 4 (-5) + 12
     • f (x) = 4 (25) + 12
     • f (x) = 112, that is, the parabola of this quadratic equation intersects the Y-axis at the point with coordinates (0, 112).
10. 10 You have found (and plotted) the vertex of the parabola, its direction, and the points of intersection with the X and Y axes. You can build parabolas from these points or find and plot additional points and only then build a parabola. To do this, plug in multiple x values ​​(on either side of the vertex) into the original equation to calculate the corresponding y values.
    • Let's return to the equation x + 2x + 1. As you already know, the point of intersection of the graph of this equation with the X-axis is the point with coordinates (-1,0). If a parabola has only one point of intersection with the X-axis, then this is the vertex of the parabola lying on the X-axis. In this case, one point is not enough to build a regular parabola. So find some extra points.
      • Let's say x = 0, x = 1, x = -2, x = -3.
      • x = 0: f (x) = (0) + 2 (0) + 1 = 1. Point coordinates: (0,1).
      • x = 1: f (x) = (1) + 2 (1) + 1 = 4. Point coordinates: (1,4).
      • x = -2: f (x) = (-2) + 2 (-2) + 1 = 1. Point coordinates: (-2,1).
      • x = -3: f (x) = (-3) + 2 (-3) + 1 = 4. Point coordinates: (-3,4).
      • Draw these points on the coordinate plane and draw a parabola (connect the points with a U-curve). Please note that the parabola is absolutely symmetrical - any point on one branch of the parabola can be mirrored (relative to the axis of symmetry) on the other branch of the parabola. This will save you time, since you do not need to calculate the coordinates of the points on both branches of the parabola.

Tips

• Round off fractional numbers (if this is a teacher's requirement) - this is how you build a correct parabola.
• If in f (x) = ax + bx + c the coefficients b or c are equal to zero, then there are no terms with these coefficients in the equation.For example, 12x + 0x + 6 becomes 12x + 6 because 0x is 0.