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1.

The following figure shows the graph of `f(x)= ax^(2)+ bx+c`, find the sign of `a, b and c`.

Answer» Fromt the graph, `f(0)= c gt 0` (as the graph intersects the y-axis on the positive side)
Also the graph is concave downwards, so `a lt0`.
Further, the vertex lies on the right-hand side of y-axis, so the abscissa of the vertex is `-(b)/(2a) gt 0 rArr b gt ` (as ` alt0`).
Discussion
2.

The graph of `(y-x)` against `(y+x)` is shown below. Which one of the following shows the graph of `y` against `x` ?

Answer» From the graph of `(y-x)` against `(y+x)`, we have
`" "(y-x)/(y+x)= k (k gt 1)`
`rArr" "y-x=k(y+x)`
`rArr" "y(1-k)=x(1+k)`
`rArr" "y= ((1+k)/(1-k))x`
Now `" "(1+k)/(1-k)lt -1`
Hence the graph is the one which has slope less then -1, and the angle of the line is less than `135^(@)` with the positive `x`-axis.
Hence, the correct option is (c).
Discussion
3.

The entire graph of the equation `y=x^2+k x-x+9`in strictly above the `x-a xi s`if and only if`k

Answer» `y=x^(2) + (k-1)x+9= (x+ (k-1)/(2))^(2) + 9 - ((k-1)/(2))^(2)`
For the entire graph to be above the x-axis, we should have
`" "9- ((k-1)/(2))^(2) gt 0`
`rArr" "k^(2)-2k -35 lt 0 rArr (k=7)(k+5) lt 0`
`rArr -5 lt k lt 7`
Discussion