(a) Simplify \(\frac{x + 2}{x - 2} - \frac{x + 3}{x - 1}\) (b) The graph of the equation \(y = Ax^{2} + Bx + C\) passes through the point (0, 0), (1, 4) and...
Question 1 Report
(a) Simplify \(\frac{x + 2}{x - 2} - \frac{x + 3}{x - 1}\)
(b) The graph of the equation \(y = Ax^{2} + Bx + C\) passes through the point (0, 0), (1, 4) and (2, 10). Find the :
(i) value of C ; (ii) values of A and B ; (iii) co-ordinates of the other point where the graph cuts the x- axis.
Answer Details
(a) Simplifying the algebraic fractions
Use the common denominator \((x-2)(x-1)\):
\[ \frac{x+2}{x-2}-\frac{x+3}{x-1} = \frac{(x+2)(x-1)-(x+3)(x-2)}{(x-2)(x-1)}. \] \[ = \frac{(x^2+x-2)-(x^2+x-6)}{(x-2)(x-1)} = \frac{x^2+x-2-x^2-x+6}{(x-2)(x-1)}. \] \[ \boxed{\frac{4}{(x-2)(x-1)}} \]This is valid for \(x\ne2\) and \(x\ne1\), since these values would make an original denominator zero.
(b) Finding the quadratic equation
The equation is \(y=Ax^2+Bx+C\). Since the graph passes through \((0,0)\), substitute \(x=0\) and \(y=0\):
\[ 0=A(0)^2+B(0)+C, \] \[ \boxed{C=0}. \]So the equation becomes:
\[ y=Ax^2+Bx. \]Using \((1,4)\):
\[ 4=A(1)^2+B(1), \] \[ A+B=4. \]Using \((2,10)\):
\[ 10=A(2)^2+B(2), \] \[ 4A+2B=10, \] \[ 2A+B=5. \]Subtract \(A+B=4\) from \(2A+B=5\):
\[ A=1. \]Then:
\[ 1+B=4 \quad\Rightarrow\quad B=3. \]Therefore:
\[ \boxed{A=1,\quad B=3} \]and the equation of the graph is:
\[ \boxed{y=x^2+3x}. \]Other x-intercept
At an x-intercept, \(y=0\):
\[ x^2+3x=0, \] \[ x(x+3)=0. \] \[ x=0 \quad\text{or}\quad x=-3. \]\((0,0)\) is the given intercept, so the other point where the graph cuts the x-axis is:
\[ \boxed{(-3,0)}. \]Examination reminder: \((1,4)\) and \((2,10)\) are points on the graph, not roots. A root is an x-value for which \(y=0\). Factoring \(x^2+3x\) gives \(x=0\) and \(x=-3\), not \(x=3\).
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