# How to Sketch a Graph of a Parabola

By Kristy Wedel
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There are two main forms of parabola equations, the standard form and the vertex form. The standard form is y = ax^2 + bx + c and vertex form is y = a(x - h)^2 + k. An important point on a parabola is the vertex, which is either the highest point or the lowest point depending on the direction the parabola opens up. Another important feature of a parabola is the axis of symmetry, which is an imaginary line that runs down the center of the parabola and divides the function in half. You can deduce many things from the equation of a parabola just by glancing at it. This is useful in making sketching the parabola easier.

Find the vertex of the parabola. In standard form, the vertex is (x, y) where x = -b/(2a) and y = a(-b/(2a))^2 + b(-b/(2a)) + c. For example, to find the vertex of y = x^2 -- x -- 6, determine what a, b and c are. For this equation, a = 1, b = -1 and c = -6. So, x = -(-1)/(2*1) = 1/2. Plugging this back into the equation gives y = -25/4. So the vertex is at (1/2, -25/4). In vertex form, (h, k) is the vertex. Therefore, the vertex of the parabola y = (x -- 2)^2 + 3 is (2, 3) because it is in vertex form and h = 2 and k = 3.

Determine if the parabola opens upward or downward from the vertex. In the standard and vertex forms, the value of "a" determines whether the parabola opens upward or downward. If "a" is positive, then it opens up. If "a" is negative, it opens down. In the standard form example, "a" is 1, which is positive, so the parabola opens up. In the vertex form example, "a" is also 1, and it opens upward.

Find the y-intercept. To find the y-intercepts in both the standard and vertex forms, plug x = 0 into the equation. For the equation y = x^2 -- x -- 6, the y-intercept is y = 0^2 -- 0 -- 6. Therefore, y = -6. For the equation y = (x -- 2)^2 + 3, the y-intercept is y = (0 -- 2)^2 + 3 = 7.

Find the x-intercepts. This is done by plugging y = 0 into the equation. It is possible that the parabola does not have any x-intercepts. In the standard form example, y = x^2 -- x -- 6 factors into y = (x - 3)(x + 2). Set the equation equal to zero. 0 = (x - 3)(x + 2) gives the x-intercepts x = 3 and x = -2. In the vertex form example, y = (x -- 2)^2 + 3 does not have any real solutions because it does not cross the x-axis.

Connect the known points and sketch the rest of the graph using the axis of symmetry. The axis of symmetry for standard form is x = -b/(2a) is and axis of symmetry of the parabola in vertex form is x=h. If you were to fold the parabola on top of itself using the axis of symmetry, the points on either side of it would line up.