SAT Function Transformations and Vertex Form Practice Quiz

Transform \(r(x)\) to vertex form as \(2(x - 2)^2 + 2\). The minimum of \(r(x)\) is at \(x = 2\) with a value of \(2\). For \(s(x) = r(2x - 1)\), set \(2x - 1 = 2\), giving \(x = 1.5\). The minimum value remains \(2\).

Write \(y(x)\) in vertex form as \(-(x - 3)^2 + 1\). The maximum of \(y(x)\) is at \(x = 3\), which translates to \(x = 2\) for \(z(x)\).

Convert \(e(x)\) to vertex form as \(4(x - 2)^2 - 1\). The minimum of \(e(x)\) is at \(x = 2\), which translates to \(x = 5\) for \(f(x)\).

Transform \(t(x)\) to vertex form as \(-2(x - 2)^2 + 5\). The maximum of \(t(x)\) is at \(x = 2\), which translates to \(x = -2\) for \(u(x)\).

Transform \(p(x)\) into vertex form as \(2(x - 2)^2 + 2\). The minimum of \(p(x)\) is at \(x = 2\), which moves to \(x = 4\) for \(q(x)\).

Convert \(v(x)\) to vertex form as \(4(x - 2)^2 + 1\). The minimum of \(v(x)\) is at \(x = 2\), which moves to \(x = 3\) for \(w(x)\).

Express \(m(x)\) as \(-(x - 3)^2\). The maximum of \(m(x)\) is at \(x = 3\), which translates to \(x = 0\) for \(n(x)\).

Transform \(p(x)\) to vertex form as \(-3(x - 2)^2 + 7\). The maximum of \(p(x)\) is at \(x = 2\) with a value of \(7\). For \(q(x) = p(-x + 2)\), set \(-x + 2 = 2\), giving \(x = 0\). The maximum value remains \(7\).

Transform \(h(x)\) to vertex form as \((x - 3)^2 + 1\). The minimum of \(h(x)\) is at \(x = 3\). For \(j(x) = h(2x - 3)\), set \(2x - 3 = 3\), giving \(x = 3\). Solving for \(x\), we get \(x = 3\).

Transform \(t(x)\) to vertex form as \(-(x - 2)^2 + 1\). The maximum of \(t(x)\) is at \(x = 2\) with a value of \(1\). For \(u(x) = t(3x - 2)\), set \(3x - 2 = 2\), giving \(x = rac{4}{3}\). The maximum value remains \(1\).

Express \(c(x)\) in vertex form as \(-3(x - 2)^2 + 1\). The maximum of \(c(x)\) is at \(x = 2\), which translates to \(x = 1\) for \(d(x)\).

Write \(r(x)\) in vertex form as \(3(x - 2)^2 - 5\). The minimum of \(r(x)\) is at \(x = 2\), which moves to \(x = 4\) for \(s(x)\).

First, transform \(f(x)\) to vertex form as \(-2(x - 2)^2 + 5\). The maximum of \(f(x)\) is at \(x = 2\) with a value of \(5\). For \(g(x) = f(x - 4)\), the maximum occurs at \(x = 6\) with the same value of \(5\).

Transform \(a(x)\) to vertex form as \(2(x - 2)^2 + 2\). The minimum of \(a(x)\) is at \(x = 2\), which translates to \(x = 4\) for \(b(x)\).

Rewrite \(h(x)\) in vertex form as \((x - 2)^2 + 1\). The minimum of \(h(x)\) is at \(x = 2\), which translates to \(x = -1\) for \(k(x)\), but the minimum value remains \(1\).