### Functions and graphs Examples

• [GraphMissing Latex_functions and graphs02 Q20]
• [GraphMissing Latex_functions and graphs02 Q20]
• In the figure, the graph of $$y = a{(x - h)^2} + k$$ cuts the x-axis at A(-1, 0) and B, and cuts the y-axis at C. V(2, -9) is the vertex of the graph. The axis of symmetry of the graph cuts the x-axis at M.
(a) Find the coordinates of M.
(b) Find the coordinates of B.
(c) Find the area of OCVB.
• (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 mid-point 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 $$y = a{(x - 2)^2} - 9$$.

• It 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.
• $$\begin{array}{1}\frac{y}{2} = 2{x^2} - bx + 3b - 16\\y = 4{x^2} - 2bx + 6b - {32^{}}\end{array}$$
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 $$b = \underline{\underline 8}$$

• It is given that the x-coordinate and y-coordinate 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.
• Let 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 $$k = 4$$,
$$\begin{array}{1}y = 2{(x - 4)^2} + 4\\ = 2({x^2} - 8x + 16) + 4\\ = 2{x^2} - 16x + 36\end{array}$$
∴ $$b = \underline{\underline { - 16}} {\rm{ }},{\rm{ }}c = \underline{\underline {36}}$$