How do you find the vertex of #y=x^2 + 4x + 2#?
Convert the given equation into vertex form to find:
vertex at
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To find the vertex of the quadratic function ( y = x^2 + 4x + 2 ), follow these steps:
- Identify the coefficients ( a ), ( b ), and ( c ) in the quadratic equation ( y = ax^2 + bx + c ).
- Use the formula ( x = \frac{{-b}}{{2a}} ) to find the x-coordinate of the vertex.
- Substitute the x-coordinate obtained in step 2 into the original equation to find the y-coordinate of the vertex.
- The coordinates of the vertex are ((x, y)).
Using the given equation ( y = x^2 + 4x + 2 ):
- ( a = 1 ), ( b = 4 ), and ( c = 2 ).
- ( x = \frac{{-4}}{{2(1)}} = -2 ).
- Substitute ( x = -2 ) into the equation: ( y = (-2)^2 + 4(-2) + 2 = 4 - 8 + 2 = -2 ).
- The vertex is at ((-2, -2)).
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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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- What is the vertex of # y = 1/2 (3x + 2)^2 - 1/3 #?
- How do you find the vertex and intercepts for #y = x^2 - 2#?
- How do you write a quadratic equation that passes through (0,9) and vertex is (-2,5)?
- How do you factor and solve #x^2 - x - 20 = 0#?

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