Find the vertex of the following parabola. please write your answer in the form: (x,y) with no spaces. (So a vertex of 2,1 would
be written as (2,1)).
Show all your work.
Show all your work.
1 answer:
Answer:
Vertex (-3,1)
Step-by-step explanation:
y = x^2 + 6x + 10 Put brackets around the first two terms
y = (x^2 + 6x) + 10 Take 1/2 the linear term (6) and square it. Put that inside the brackets. Always add.
y = (x^2 + 6x + (6/2)^2 ) + 10 Since you added in side the brackets, subtract after the 10
y = (x^2 + 6x + 9) + 10 - 9 The 3 terms inside the brackets are a perfect square. Combine the two terms outside the brackets.
y = (x + 3)^2 + 1
The vertex is at (the value that makes x + 3 go to zero, and the number outside the brackets)
(-3,1)
Check
The check is to ask Desmos to graph the equation. That graph is shown below. Notice that the lowest point is(-3,1). Notice that the graph and my answer do agree.
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