Question

The vertex form of the equation of a parabola is y=(x-5)2+16.What is the standard form of the equation?
a. y=x2-10x+41
b. y=x2+10x+16
c. y=x2+x+8
d. y=3x2-10x+41

Answer 1

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

The answer is A.

Answer 2

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 $y=x^2-10x+41$