Stitz-Zeager_College_Algebra_e-book

We can see this analytically by substituting g x x2 x

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: f 2a − b 2a . Example 2.3.2. Use Equation 2.4 to find the vertex of the graphs in Example 2.3.1. Solution. 1. The formula f (x) = x2 − 4x + 3 is in the form f (x) = ax2 + bx + c. We identify a = 1, b = −4, and c = 3, so that b −4 − =− = 2, 2a 2(1) and f − b 2a = f (2) = −1, so the vertex is (2, −1) as previously stated. 1 We will justify the role of a in the behavior of the parabola later in the section. 142 Linear and Quadratic Functions 2. We see that the formula g (x) = −2(x − 3)2 + 1 is in the form g (x) = a(x − h)2 + k . We identify a = −2, x − h as x − 3 (so h = 3), and k = 1 and get the vertex (3, 1), as required. The formula f (x) = a(x − h)2 + k , a = 0 in Equation 2.4 is sometimes called the standard form of a quadratic function; the formula f (x) = ax2 + bx + c, a = 0 is sometimes called the general form of a quadratic function. To see why the formulas in Equation 2.4 produce the vertex, let us first consider a quadratic function in standard form. If we consider the graph of the equation y = a(x − h)2 + k we see that when x...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online