What are the coordinates of the vertex of y=x²+4x+3?

Mathematics

1 answer:

Travka [436]3 years ago

4 0

Answer:

(-2, -1).

Step-by-step explanation:

y = x² + 4x + 3

Completing the square:

y = (x + 4/2)^2 - (4/2)^2 + 3

y = (x + 2)^2 - 4 + 3

y = (x + 2)^2 - 1.

This is vertex form and the coordinates of the vertex are (-2, -1).

You might be interested in

Can someone please help me with these 2 questions

soldi70 [24.7K]

Number 8 would be B. Number 9 would be A. Hope this helps! Mark brainliest? :))

3 0

3 years ago

Read 2 more answers

6. A government's economic role is complicated by the fact that

user100 [1]

Answer:B

Step-by-step explanation:

5 0

3 years ago

The answer is red show step by step of the pro tem to get the answer

miskamm [114]

In order to rationalize the denominator of each expression, we need to multiply the expression by the same radical in the denominator, this way we can remove the radical from the denominator.

$\frac{5\sqrt{4}}{\sqrt{3}}(\cdot\sqrt{3})=\frac{5\sqrt{4}\sqrt{3}}{(\sqrt{3})^2}=\frac{5\cdot2\cdot\sqrt{3}}{3}=\frac{10\sqrt{3}}{3}$

10)

$-\frac{5}{3\sqrt{2}}(\cdot\sqrt{2})=-\frac{5\sqrt{2}}{3(\sqrt{2})^2}=-\frac{5\sqrt{2}}{3\cdot2}=-\frac{5\sqrt{2}}{6}$

11)

$\frac{2\sqrt{3}}{4\sqrt{5}}(\cdot\sqrt{5})=\frac{2\sqrt{3}\sqrt{5}}{4(\sqrt{5})^2}=\frac{2\sqrt{15}}{4\cdot5}=\frac{\sqrt{15}}{2\cdot5}=\frac{\sqrt{15}}{10}$

12)