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How do you write #f(x) = x^2 - 3x + 1 # in vertex form and identify the vertex?

Answer 1

Savannah Anderson

x o vertex: #x = (-b/2a) = 3/2#

y of vertex:# f(3/2) = 9/4 - 9/2 + 1 = -5/4#

Vertex (3/2, -5/4)

Vertex form:# f(x) = (x - 3/2)^2 - 5/4#

Check by developing:# f(x) = x^2 - 3x + 9/4 - 5/4 = x^2 - 3x + 1# .OK

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Answer 2

Eva Abramson

To write ( f(x) = x^2 - 3x + 1 ) in vertex form, we complete the square. First, factor out the leading coefficient of ( x^2 ), which is 1. Then, add and subtract ((\frac{-3}{2})^2 = \frac{9}{4}) inside the parentheses:

( f(x) = x^2 - 3x + 1 )
( = (x^2 - 3x + \frac{9}{4}) - \frac{9}{4} + 1 )
( = (x^2 - 3x + \frac{9}{4}) - \frac{5}{4} )
( = (x - \frac{3}{2})^2 - \frac{5}{4} )

The vertex form is ( f(x) = a(x - h)^2 + k ), where ( (h, k) ) is the vertex. So, the vertex of the parabola is ( (h, k) = (\frac{3}{2}, -\frac{5}{4}) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.