How do you write #4x^2+16-9# in vertex form and identify the vertex, y intercept and x intercept?

Answer 1

Please check your question. f(x) = 4x^2 + 16x - 9 ?

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Answer 2

To write ( 4x^2 + 16x - 9 ) in vertex form, complete the square.

[ 4x^2 + 16x - 9 = 4(x^2 + 4x) - 9 ] [ = 4(x^2 + 4x + 4 - 4) - 9 ] [ = 4((x + 2)^2 - 4) - 9 ] [ = 4(x + 2)^2 - 16 - 9 ] [ = 4(x + 2)^2 - 25 ]

The vertex form is ( y = 4(x + 2)^2 - 25 ).

To identify the vertex, y-intercept, and x-intercept:

Vertex: The vertex is (-2, -25).

Y-intercept: Set ( x = 0 ) and solve for ( y ). [ y = 4(0 + 2)^2 - 25 ] [ y = 4(4) - 25 ] [ y = 16 - 25 ] [ y = -9 ]

The y-intercept is (0, -9).

X-intercept: Set ( y = 0 ) and solve for ( x ). [ 4(x + 2)^2 - 25 = 0 ] [ 4(x + 2)^2 = 25 ] [ (x + 2)^2 = \frac{25}{4} ] [ x + 2 = \pm \frac{5}{2} ] [ x = -2 \pm \frac{5}{2} ]

The x-intercepts are ( x = -2 + \frac{5}{2} ) and ( x = -2 - \frac{5}{2} ). Simplify to find the exact values.

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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.

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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