Tangent to curve: find equation

A question is this type if and only if it asks to find the equation of a tangent line to a quadratic curve, or coordinates where tangency occurs.

2 questions · Moderate -0.8

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OCR MEI C1 Q3
12 marks Moderate -0.8
  1. Express \(x^2 - 6x + 2\) in the form \((x - a)^2 - b\). [3]
  2. State the coordinates of the turning point on the graph of \(y = x^2 - 6x + 2\). [2]
  3. Sketch the graph of \(y = x^2 - 6x + 2\). You need not state the coordinates of the points where the graph intersects the \(x\)-axis. [2]
  4. Solve the simultaneous equations \(y = x^2 - 6x + 2\) and \(y = 2x - 14\). Hence show that the line \(y = 2x - 14\) is a tangent to the curve \(y = x^2 - 6x + 2\). [5]
OCR MEI C1 Q6
13 marks Moderate -0.8
\includegraphics{figure_6} Fig. 11 shows a sketch of the curve with equation \(y = (x - 4)^2 - 3\).
  1. Write down the equation of the line of symmetry of the curve and the coordinates of the minimum point. [2]
  2. Find the coordinates of the points of intersection of the curve with the \(x\)-axis and the \(y\)-axis, using surds where necessary. [4]
  3. The curve is translated by \(\begin{pmatrix} 2 \\ 0 \end{pmatrix}\). Show that the equation of the translated curve may be written as \(y = x^2 - 12x + 33\). [2]
  4. Show that the line \(y = 8 - 2x\) meets the curve \(y = x^2 - 12x + 33\) at just one point, and find the coordinates of this point. [5]