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Find analytical expression of a quadratic function example2

Question:

The graph of a quadratic function, y = ax2 + bx + c, has the vertex A (-1, 3) and passes the point B (1, -5). Find the analytical expression of the graph.

find analytical expression of the quadratic function y = ax^2 + bx + c when given vertex and other point.

Solution:

Because the point A lies on the graph and the point A is the vertex of the graph, so, the function of the graph is: y = a (x + m)2 + k, in which (-m, k) is the vertex. That is, -m is the x-coordinate of the vertex and k is y-coordinate of the vertex.

Substitute the vertex coordinate A (-1, 3) into the quadratic function, we get, y = a (x + 1)2 + 3, in which the vertex of this graph is (-1, 3). Now we need to determine the coefficient a of the x square term.

Because the point B lies on the graph, so, the point B satisfy the quadratic function. Substitute the coordinate of the point B (1, -5) into the quadratic function y = a (x + 1)2 + 3, we get

-5 = a (1 + 1)2 + 3
-5 = a × 22 + 3
-5 = 4a + 3
4a = -5 - 3
4a = -8
a = -2

Therefore, the quadratic function of the graph is: y = -2 (x + 1)2 + 3. Now we write it as general form

y = -2 (x + 1)2 + 3
= -2 (x2 + 2x + 1) + 3
= -2x2 - 4x - 2 + 3
= -2x2 - 4x + 1

Therefore, the quadratic function of the graph is y = -2x2 - 4x + 1. Watch the video for more details.