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The vertex form of the equation of a parabola is y=(x-5)2+16.What is the standard form of the equation?

Ay=x2-10x+41
By=x2+10x+16
Cy=x2+x+8
Dy=3x2-10x+41

Verified Answer :

naǫ

\(y=(x-5)^2+16 \\ y=x^2-2 \times x \times 5+5^2+16 \\ y=x^2-10x+25+16 \\ y=x^2-10x+41\)

The answer is A.

Answer:

Option (A) is correct.

The standard form of the equation of parabola  \(y=(x-5)^2+16\)  is \(y=x^2-10x+41\)

Step-by-step explanation:

Given : The equation of a parabola is \(y=(x-5)^2+16\)

We have to find the standard form of the equation of parabola

Consider the equation of a parabola is \(y=(x-5)^2+16\)

Applying algebraic identity \((a-b)^2=a^2+b^2-2ab\)

We have, a= x and b = 5

\((x-5)^2=x^2+5^2-2\cdot x\cdot 5\)

Simplify, we have,

\((x-5)^2=x^2-10x+25\)

Substitue, we have,

\(y=x^2-10x+25+16\)

Simplify, we have,

\(y=x^2-10x+41\)

Thus, The standard form of the equation of parabola  \(y=(x-5)^2+16\)  is \(y=x^2-10x+41\)

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