Quick Check Solutions Graphing Quadratics Target C
1) The \(y\)-intercept of \(f(x)\) is \((0, -5)\) - we can find this by substituting in \(x = 0\), the \(y\)-intercept of \(g(x)\) is \((0, 7)\) we find where the graph crosses the \(y\)-axis and the \(y\)-intercept of \(h(x)\) is \((0, 5)\) we find this by looking at the table for \(x = 0\). So the \(y\)-intercept that is the lowest in value would be that of \(f(x)\).
2) \(g(x)\) and \(h(x)\) have minimum values because both are opening up.
3) \(f(1) = 0\), we find this by substituting into the equation.
\(g(1) = 4\) we find this by finding the \(y\)-value when \(x = 1\)
\(h(1) = 2\) we find this by finding \(x = 1\) in the table and reading the y-value associated with it so \(f(1) < h(1) < g(1)\).
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1) The \(y\)-intercept of \(f(x)\) is \((0, -5)\) - we can find this by substituting in \(x = 0\), the \(y\)-intercept of \(g(x)\) is \((0, 7)\) we find where the graph crosses the \(y\)-axis and the \(y\)-intercept of \(h(x)\) is \((0, 5)\) we find this by looking at the table for \(x = 0\). So the \(y\)-intercept that is the lowest in value would be that of \(f(x)\).
2) \(g(x)\) and \(h(x)\) have minimum values because both are opening up.
3) \(f(1) = 0\), we find this by substituting into the equation.
\(g(1) = 4\) we find this by finding the \(y\)-value when \(x = 1\)
\(h(1) = 2\) we find this by finding \(x = 1\) in the table and reading the y-value associated with it so \(f(1) < h(1) < g(1)\).
Back to Guided Learning