Question 105601
Question 105601
Free
Express \(y={{x}^{2}}-8x+17\) in the form of \(y={{(x-h)}^{2}}+k\). Hence, state
- the maximum or minimum point of the function and the value of \(x\) where it occurs
- the equation of the line of symmetry of the graph.
Sketch the graph and indicate clearly the turning point and cutting point(s) on the axes.
Default solution
\(y\)\({}=x^2-8x+17\)\({}=(x^2-8x+16)-16+17\)\({}=(x-4)^2+1\)
\(h=4,\qquad k=1\)
- \((x-4)^2\ge0\), so the minimum point is \((4,1)\), occurring at \(x=4\).
- Line of symmetry: \(x=4\).
At \(x=0\), \(y=17\), so the \(y\)-intercept is \((0,17)\).
\(y=(x-4)^2+1\ge1\), so there are no \(x\)-intercepts.
Answer:\(y=(x-4)^2+1\) (i) Minimum point \((4,1)\) at \(x=4\) (ii) \(x=4\). The \(y\)-intercept is \((0,17)\); no \(x\)-intercepts.
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