How do you write #f(x) = x^2 + 7x + 2# in vertex form?
In vertex form:
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To write the function ( f(x) = x^2 + 7x + 2 ) in vertex form, complete the square.
Start by rearranging the terms to group the ( x ) terms together:
[ f(x) = (x^2 + 7x) + 2 ]
Next, complete the square for the ( x^2 + 7x ) term by adding and subtracting ((\frac{7}{2})^2 = \frac{49}{4}):
[ f(x) = (x^2 + 7x + \frac{49}{4}) - \frac{49}{4} + 2 ]
Now, rewrite the expression inside the parentheses as a perfect square:
[ f(x) = (x + \frac{7}{2})^2 - \frac{49}{4} + 2 ]
Simplify:
[ f(x) = (x + \frac{7}{2})^2 - \frac{49}{4} + \frac{8}{4} ]
Combine the constant terms:
[ f(x) = (x + \frac{7}{2})^2 - \frac{41}{4} ]
So, the function ( f(x) = x^2 + 7x + 2 ) in vertex form is ( f(x) = (x + \frac{7}{2})^2 - \frac{41}{4} ).
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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.
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.
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.
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.
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