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What is the y-coordinate of the vertex of the parabolay=x^2+8x+17

User Lejo
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Answer:

y-coordinate = 1

Explanation:

Identifying the form of y = x^2 + 8x + 17:

The equation y = x^2 + 8x + 17 is in the standard form of a quadratic, whose general equation is given by:

y = ax^2 + bx + c, ,where

  • a, b, and c are constants.

Thus, for y = x^2 + 8x + 17, 1 is our a value, 8 is our b value, and 17 is our c value.

Using this information to find the y-coordinate of the vertex:

We can first find the x-coordinate of the vertex using the formula -b / 2a.

Thus, we can find the x-coordinate of the vertex by substituting 8 for b and 1 for a:

x-coordinate = -8 / 2(1)

x-coordinate = -8 / 2

x-coordinate = -4

Thus, the x-coordinate of the vertex is -4.

Now we can find the y-coordinate of the vertex by substituting -4 for x in y = x^2 + 8x + 17 and solving for y:

y = (-4)^2 + 8(-4) + 17

y = 16 - 32 + 17

y = -16 + 17

y = 1

Thus, the y-coordinate of the vertex is 1.

User Lashana
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