Functions and graphs Examples

Graph[GraphMissing Latex_functions and graphs02 Q20]

Theory[GraphMissing Latex_functions and graphs02 Q20]

ExamplesIn the figure, the graph of \(y = a{(x  h)^2} + k\) cuts the xaxis at A(1, 0) and B, and cuts the yaxis at C. V(2, 9) is the vertex of the graph. The axis of symmetry of the graph cuts the xaxis at M.
(a) Find the coordinates of M.
(b) Find the coordinates of B.
(c) Find the area of OCVB. 
Solutions(a) Let the coordinates of M be \(({x_1}{\rm{ }},{\rm{ }}0)\).
∵ The axis of symmetry of the graph passes through the vertex (2, 9).
∴ \({x_1} = 2\)
∴ The coordinates of M are (2, 0).
(b) Let the coordinates of B be \(({x_2}{\rm{ }},{\rm{ }}0)\).
∵ M is the midpoint of A and B.
∴ \(\begin{array}{c}2 = \frac{{{x_2} + (  1)}}{2}\\{x_2}  1 = 4\\{x_2} = 5\end{array}\)
∴ The coordinates of B are (5, 0).
(c) ∵ The coordinates of the vertex are (2, 9).
∴ The function can be expressed as \(y = a{(x  2)^2}  9\).

Graph

Theory

ExamplesIt is given that \(\frac{y}{2} = 2{x^2}  bx + 3b  16\), where b is a constant, If the minimum value of y is zero, find the values of b.

Solutions\(\begin{array}{1}\frac{y}{2} = 2{x^2}  bx + 3b  16\\y = 4{x^2}  2bx + 6b  {32^{}}\end{array}\)
The minimum value of y = 0
\(\begin{array}{1}\frac{{4(4)(6b  32)  {{(  2b)}^2}}}{{4(4)}} = 0\\96b  512  4{b^2} = 0\\{b^2}  24b + 128 = 0\\(b  16)(b  8) = 0\end{array}\)
\(b  16 = 0\) or \(b  8 = 0\)
\(b = \underline{\underline {16}} \) or \(b = \underline{\underline 8} \)

Graph

Theory

ExamplesIt is given that the xcoordinate and ycoordinate of the vertex of the graph of \(y = 2{x^2} + bx + c\) are equal, where b and c are constants. If P(2, 12) is a point on the graph, find the values of b and c.

SolutionsLet the coordinates of the vertex be (k, k) .
∴ The function can be expressed as.\(y = 2{(x  k)^2} + k\).
∵ The graph passes through P(2, 12).
\(\begin{array}{1}∴12 = 2{(2  k)^2} + k\\12 = (8  8k + 2{k^2}) + k\\2{k^2}  7k  4 = 0\\(2k + 1)(k  4) = 0\end{array}\)
\(2k + 1 = 0\) or \(k  4 = 0\)
\(k =  \frac{1}{2}\) or \(k = 4\)
When \(k =  \frac{1}{2}\),
\(\begin{array}{1}y = 2{(x + \frac{1}{2})^2}  \frac{1}{2}\\ = 2({x^2} + x + \frac{1}{4})  \frac{1}{2}\\ = 2{x^2} + 2x\end{array}\)
∴ \(b = \underline{\underline 2} {\rm{ }},{\rm{ }}c = \underline{\underline 0} \)
When \(k = 4\),
\(\begin{array}{1}y = 2{(x  4)^2} + 4\\ = 2({x^2}  8x + 16) + 4\\ = 2{x^2}  16x + 36\end{array}\)
∴ \(b = \underline{\underline {  16}} {\rm{ }},{\rm{ }}c = \underline{\underline {36}} \)