# Vertex and Intercepts Parabola Problems

A set of problems, with detailed solutions, related to the vertex and intercepts of parabolas are presented.

 Review The graph of a quadratic function of the form f(x) = a x 2 + b x + c is a parabola with vertex at the point (h , k). where h = - b / 2a and and k = f(h) = c - b 2 / 4a Also the x intercepts (if any) of the graph of f are given by x1 = [ - b + sqrt (b 2 - 4 a c) ] / 2 a and x2 = [ - b - sqrt (b 2 - 4 a c) ] / 2 a It can be easily shown that h = (x1 + x2) / 2 If the graph of f has x intercepts at x1 and x2 then x1 and x2 are solutions to the equation f(x) = 0 so that f may be given by f(x) = a (x - x1) (x - x2) Problem 1 : Find the quadratic function whose graph, which is a parabola, has x intercepts at x = - 4 and x = 6, and its highest point has a y coordinate equal to 5. Solution to Problem 1: The x intercepts are given as - 4 and 6 hence f may be written as f(x) = a (x - (-4)) (x - 6) The highest point is the vertex. If x1 and x2 are the x intrcepts of the graph then the x coordinate h of the vertex is given by h = (x1 + x2) / 2 = (- 4 + 6) / 2 = 1 We now know the x (h = 1) and y coordinates (k = 6) of the vertex which is a point on the graph of the parabola. Hence f( 1 ) = 6 When f( 1 ) = 6 is substituted in f(x) = a (x - (-4)) (x - 6), it gives a (1 + 4)(1 - 6) = 6 Solve for a to find a = - 6 / 25 The equation of the parabola is given by f(x) = ( - 6 / 25) (x + 4) (x - 6) Problem 2 : A parabola, with verical axis, has a vertex at (1 , - 8) and passes through the point (2 , - 6). Find the x intercepts of this parabola. Solution to Problem 2: If h and k are the coordinates of the vertex, the quadratic function in vertex form corresponding to this parabola is given by f(x) = a (x - h) 2 + k = a (x - 1) 2 - 8 Use the fact that (2 , - 6) is a point on the graph of the parabola to find coefficient a - 6 = a (2 - 1) 2 - 8 Solve for a to obtain a = 2 The equation of the parabola is given by f(x) = 2 (x - 1) 2 - 8 The x intercepts of the parabola are found by solving f(x) = 0 Hence 2 (x - 1) 2 - 8 = 0 Solve for x 2 (x - 1) 2 - 8 = 0 2 (x - 1) 2 = 8 (x - 1) 2 = 4 solutions to above equation are x = 3 and x = -1 The x intercepts are at the points : (3 , 0) and (-1 , 0). Problem 3 : A quadratic function has a minimum value of - 2 and its graph has y intercept at (0 , 6) and x intercept at (3 , 0). Find a possible equation for this quadratic function. Solution to Problem 2: The minimum value of a quadratic function is equal to k in the quadratic function in vertex form which is given by f(x) = a (x - h) 2 + k = a (x - h) 2 - 2 We now use the y intercept (0 , 6) given above 6 = a (0 - h) 2 - 2 = a h 2 - 2 Solve the above equation for a a = 8 / h 2 Use the x intercept (3 , 0) given in the problem above 0 = a (3 - h) 2 - 2 Substitute a = 8 / h 2 into the above equation to obtain 0 = ( 8 / h 2 )(3 - h) 2 - 2 Eliminate the denominators to obtain a quadratic equation in h h 2 - 8 h + 12 = 0 Solve the above equation for h h = 6 and h = 2. Use a = 8 / h 2 to find the corresponding values of a when h = 6 , a = 2 / 9 and when h = 2 , a = 2 The two quadratic functions that are solutions are given by f(x) = 2 (x - 2) 2 - 2 f(x) = (2 / 9) (x - 6) 2 - 2 The graphs (parabolas) of the two quadratic functions obtained are shown below. Check that they have the same x , y intrcepts and minimum value. More Online Tutorials on Functions and Algebra.