Question

station 4: schoology staar practice\
quadratic function f can be written as $f(x)=(x - 7)^2 + 1$ and $f(x)=x^2 + kx + 50$. what is the value of k?\
k = -54

Explanation:

Step1: Expand the vertex form

We know that \( f(x)=(x - 7)^2+1 \). Using the formula \((a - b)^2=a^2-2ab + b^2\), where \(a = x\) and \(b = 7\), we expand \((x - 7)^2\) as \(x^2-14x + 49\). Then \(f(x)=x^2-14x + 49+1\).

Step2: Simplify the expanded form

Simplify \(f(x)=x^2-14x + 49+1\) to \(f(x)=x^2-14x + 50\).

Step3: Compare with the standard form

We are given that \(f(x)=x^2+kx + 50\). By comparing the coefficients of \(x\) in \(x^2-14x + 50\) and \(x^2+kx + 50\), we can see that \(k=-14\). Wait, but the initial box has -54, that might be a mistake. Let's re - check.

Wait, maybe I made a mistake. Wait, \((x - 7)^2=x^2-14x + 49\), then \((x - 7)^2+1=x^2-14x+49 + 1=x^2-14x + 50\). And the other form is \(x^2+kx + 50\). So when we compare \(x^2-14x + 50\) and \(x^2+kx + 50\), the coefficient of \(x\) is \(-14\) for the first one and \(k\) for the second one. So \(k=-14\). The initial -54 is incorrect.

Let's do it again:

Step1: Expand \((x - 7)^2\)

Using the perfect square formula \((a - b)^2=a^{2}-2ab + b^{2}\), with \(a = x\) and \(b = 7\), we get \((x - 7)^2=x^{2}-2\times x\times7+7^{2}=x^{2}-14x + 49\).

Step2: Add 1 to the expanded form

\(f(x)=(x - 7)^2+1=x^{2}-14x + 49+1=x^{2}-14x + 50\).

Step3: Compare with \(x^{2}+kx + 50\)

Since \(f(x)=x^{2}-14x + 50\) and \(f(x)=x^{2}+kx + 50\), the coefficient of \(x\) in both expressions must be equal. So \(k=-14\).

Answer:

\(k = - 14\)