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boyakko [2]
4 years ago
5

How do you find the axis of symmetry in the quadratic equation x²+4x+4=0

Mathematics
1 answer:
frez [133]4 years ago
7 0
The equation of a parabola is y=a(x-p)^2+q, where (p,q) are the coordinates of the vertex. The value of p will be your axis of symmetry.

(x+2)^2=0

In order for 0=0

x=-2

So your axis of symmetry is x=-2
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Cedric completed the square for the quadratic expression 5x^2−40x+15 in order to determine the minimum value of the expression,
Olegator [25]

Answer:

Step 3 is not correct (should be (x-4)^2, not (x+4)^2

Step-by-step explanation:

The expression that we have to simplify is:

5x^2-40x+15

We proceed step-by-step, in order to find the mistake did by Cedric:

1) First, we factorize the 5 outside:

5(\frac{5x^2}{5}-\frac{40x}{5}+\frac{15}{5})=5(x^2-40x+3) --> this step is correct

2) We add +16 and -16 inside the brackets:

5(x^2-8x+16-16+3) --> this step is correct

3) We rewrite the term (x^2-8x+16) as the square of a bynomial, which is (x-4)^2, so the expression becomes

5((x-4)^2-16+3) --> we notice that this step is wrong: in fact, Cedric wrote (x+4)^2, which is not correct.

4) Now we continue: we rewrite -16+3 as -13,

5((x-4)^2-13)

5) Finally, we multiply the 5 by the terms in the bracket:

=5(x-4)^2-65

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