How To Graph A Quadratic Equation 10 Steps With Pictures

Standard form. [3] X Research source In this form, the quadratic equation is written as: f(x) = ax2 + bx + c where a, b, and c are real numbers and a is not equal to zero. For example, two standard form quadratic equations are f(x) = x2 + 2x + 1 and f(x) = 9x2 + 10x -8. Vertex form. [4] X Research source In this form, the quadratic equation is written as: f(x) = a(x - h)2 + k where a, h, and k are real numbers and a does not equal zero. Vertex form is so named because h and k directly give you the vertex (central point) of your parabola at the point (h,k). Two vertex form equations are f(x) = 9(x - 4)2 + 18 and -3(x - 5)2 + 1 To graph either of these types of equations, we need to first find the vertex of the parabola, which is the central point (h,k) at the “tip” of the curve. The coordinates of the vertex in standard form are given by: h = -b/2a and k = f(h), while in vertex form, h and k are specified in the equation.

For example, two standard form quadratic equations are f(x) = x2 + 2x + 1 and f(x) = 9x2 + 10x -8.

Two vertex form equations are f(x) = 9(x - 4)2 + 18 and -3(x - 5)2 + 1

For example, for the standard form equation f(x) = 2x2 +16x + 39, we have a = 2, b = 16, and c = 39. For the vertex form equation f(x) = 4(x - 5)2 + 12, we have a = 4, h = 5, and k = 12. [6] X Research source

In our standard form example (f(x) = 2x2 +16x + 39), h = -b/2a = -16/2(2). Solving, we find that h = -4. In our vertex form example (f(x) = 4(x - 5)2 + 12), we know h = 5 without doing any math.

We have determined in our standard form example that h = -4. To find k, we solve our equation with our value for h replacing x: k = 2(-4)2 + 16(-4) + 39. k = 2(16) - 64 + 39. k = 32 - 64 + 39 = 7 In our vertex form example, again, we know the value of k (which is 12) without having to do any math.

In our standard form example, our vertex will be at (-4,7). So, our parabola will peak 4 spaces to the left of 0 and 7 spaces above (0,0). We should plot this point on our graph, being sure to label coordinates. In our vertex form example, our vertex is at (5,12). We should plot a point 5 spaces to the right and 12 spaces above (0,0).

In the case of our standard form example, the axis is a line parallel to the y-axis and passing through the point (-4, 7). Though it’s not part of the parabola itself, lightly marking this line on your graph can eventually help you see how the parabola curves symmetrically.

For our standard form example (f(x) = 2x2 +16x + 39), we know we have a parabola opening upwards because, in our equation, a = 2 (positive). For our vertex form example (f(x) = 4(x - 5)2 + 12), we know we have also have a parabola opening upwards because a = 4 (positive).

Simply set f(x) = 0 and solve the equation. This method may work for simple quadratic equations, especially in vertex form, but will prove exceedingly difficult for more complicated ones. See below for an example f(x) = 4(x - 12)2 - 4 0 = 4(x - 12)2 - 4 4 = 4(x - 12)2 1 = (x - 12)2 SqRt(1) = (x - 12) +/- 1 = x -12. x = 11 and 13 are the parabola’s x-intercepts. Factor your equation. Some equations in the ax2 + bx + c form can be easily factored into the form (dx + e)(fx +g), where dx × fx = ax2, (dx × g + fx × e) = bx, and e × g = c. In this case, your x intercepts are the values for x which make either term in parentheses = 0. For example: x2 + 2x + 1 = (x + 1)(x + 1) In this case, your only x intercept is -1 because setting x equal to -1 will make either of the factored terms in parentheses equal 0. Use the quadratic formula. [12] X Research source If you can’t easily solve for your x intercepts or factor your equation, use a special equation called the quadratic formula designed for this very purpose. If it isn’t already, get your equation into the form ax2 + bx + c, then plug a, b, and c into the formula x = (-b +/- SqRt(b2 - 4ac))/2a. [13] X Research source Note that this often gives you two answers for x, which is OK - this just means your parabola has two x intercepts. See below for an example: -5x2 + 1x + 10 gets plugged into the quadratic formula as follows: x = (-1 +/- SqRt(12 - 4(-5)(10)))/2(-5) x = (-1 +/- SqRt(1 + 200))/-10 x = (-1 +/- SqRt(201))/-10 x = (-1 +/- 14. 18)/-10 x = (13. 18/-10) and (-15. 18/-10). The parabola’s x intercepts are at approximately x = -1. 318 and 1. 518 Our previous standard form example, 2x2 + 16x + 39 gets plugged into the quadratic formula as follows: x = (-16 +/- SqRt(162 - 4(2)(39)))/2(2) x = (-16 +/- SqRt(256 - 312))/4 x = (-16 +/- SqRt(-56)/-10 Because finding the square root of a negative number is impossible, we know that no x intercepts exist for this particular parabola.

For example, we know our quadratic equation 2x2 + 16x + 39 has a y intercept at y = 39, but it can also be found as follows: f(x) = 2x2 + 16x + 39 f(x) = 2(0)2 + 16(0) + 39 f(x) = 39. The parabola’s y intercept is at y = 39. As noted above, the y intercept is at y = c. Our vertex form equation 4(x - 5)2 + 12 has a y intercept that can be found as follows: f(x) = 4(x - 5)2 + 12 f(x) = 4(0 - 5)2 + 12 f(x) = 4(-5)2 + 12 f(x) = 4(25) + 12 f(x) = 112. The parabola’s y intercept is at y = 112.

Let’s revisit the equation x2 + 2x + 1. We already know its only x intercept is at x = -1. Because it only touches the x intercept at one point, we can infer that its vertex is its x intercept, which means its vertex is (-1,0). We effectively only have one point for this parabola - not nearly enough to draw a good parabola. Let’s find a few more to ensure we draw an accurate graph. Let’s find the y values for the following x values: 0, 1, -2, and -3. For 0: f(x) = (0)2 + 2(0) + 1 = 1. Our point is (0,1). For 1: f(x) = (1)2 + 2(1) + 1 = 4. Our point is (1,4). For -2: f(x) = (-2)2 + 2(-2) + 1 = 1. Our point is (-2,1). For -3: f(x) = (-3)2 + 2(-3) + 1 = 4. Our point is (-3,4). Plot these points to the graph and draw your U-shaped curve. Note that the parabola is perfectly symmetrical - when your points on one side of the parabola lie on whole numbers, you can usually save yourself some work by simply reflecting a given point across the parabola’s axis of symmetry to find the corresponding point on the other side of the parabola.