Q. \(x^2 – 8x + 16\)

Q: Find y-intercept of y=(x-4)^2.

Substitute x=0: y=(0-4)^2=16, so (0,16)

Q: Find zeros of y=(x-4)^2.

Set (x-4)^2=0, giving x=4 (double root)

Answer

Expand the square using \((a-b)^2=a^2-2ab+b^2\) with \(a=x\) and \(b=4\).

\[
(x-4)^2=x^2-2(x)(4)+4^2=x^2-8x+16
\]

Detailed Explanation

We want to simplify the expression \( (x-4)^2 \). This is a perfect square trinomial, so we can expand it using the formula for squaring a binomial.

Step 1: Use the identity for a square of a binomial

For any expressions \(a\) and \(b\), we have the identity:

\[
(a-b)^2 = a^2 – 2ab + b^2
\]

Step 2: Identify \(a\) and \(b\) in the given problem

In \( (x-4)^2 \), we match it to \( (a-b)^2 \) by letting:

\(a = x\)
\(b = 4\)

Step 3: Substitute \(a=x\) and \(b=4\) into the formula

Substitute into \(a^2 – 2ab + b^2\):

\[
(x-4)^2 = x^2 – 2(x)(4) + 4^2
\]

Step 4: Simplify each term

Now simplify the middle term and the last term:

\(x^2\) stays as \(x^2\)
\(-2(x)(4) = -8x\)
\(4^2 = 16\)

So the expression becomes:

\[
(x-4)^2 = x^2 – 8x + 16
\]

Final Answer

\[
(x-4)^2 = x^2 – 8x + 16
\]

See full solution

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Algebra FAQ

Expand \((x-4)^2\).

\((x-4)^2=x^2-8x+16\)

What is the vertex of \(y=(x-4)^2\).

Vertex is \((4,0)\)

Solve \((x-4)^2=0\).

\(x-4=0\), so \(x=4\)

Solve \((x-4)^2=9\).

\(x-4=\pm3\), so \(x=1\) or \(x=7\)

Find \(y\)-intercept of \(y=(x-4)^2\).

Substitute \(x=0\): \(y=(0-4)^2=16\), so \((0,16)\)

Find zeros of \(y=(x-4)^2\).

Set \((x-4)^2=0\), giving \(x=4\) (double root)

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