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a classmate said that the vertex of $f(x)=-5(x + 2)^{2}-6$ is $(2,0)$. …

Question

a classmate said that the vertex of $f(x)=-5(x + 2)^{2}-6$ is $(2,0)$. what mistake did your classmate make? and what is the correct vertex?

Explanation:

Step1: Recall vertex - form of a quadratic function

The vertex - form of a quadratic function is \(y = a(x - h)^2+k\), where the vertex is \((h,k)\).
For the function \(f(x)=-5(x + 2)^2-6\), we can rewrite it as \(f(x)=-5(x-(- 2))^2-6\).

Step2: Identify the values of \(h\) and \(k\)

Here, \(a=-5\), \(h = - 2\), and \(k=-6\).
The vertex of the function \(y = a(x - h)^2 + k\) is \((h,k)\), so the vertex of \(f(x)=-5(x + 2)^2-6\) is \((-2,-6)\).
The classmate's mistake was that they incorrectly identified the \(x\) - coordinate of the vertex. They likely thought that since the expression is \((x + 2)\), the \(x\) - coordinate of the vertex is \(2\), but in the vertex - form \(y=a(x - h)^2+k\), when we have \((x+2)=(x-(-2))\), the \(x\) - coordinate of the vertex is \(-2\), not \(2\), and they also misidentified the \(y\) - coordinate as \(0\) instead of \(-6\).

Answer:

The classmate misinterpreted the \(x\) - coordinate of the vertex from \((x + 2)\) as \(2\) instead of \(-2\) and misidentified the \(y\) - coordinate as \(0\) instead of \(-6\). The correct vertex is \((-2,-6)\).