Given is the graph of the relation \(y = ax^{2} + bx + c\) where a, b and c are constants. Use the graph to : (a) find the roots of the equation \(ax^{2} + ...

Answer Details

The curve is a downward-opening parabola, so the coefficient of \(x^2\) is negative. Its roots are the \(x\)-coordinates where the curve crosses the horizontal axis.

(a) The roots are the \(x\)-intercepts:

\[x=-1.5 \quad \text{or} \quad x=2.\]

(b) The graph crosses the \(y\)-axis at \(6\), so \(c=6\). Therefore,

\[y=ax^2+bx+6.\]

Use each root, where \(y=0\).

When \(x=2\):

\[4a+2b+6=0 \quad\Rightarrow\quad 4a+2b=-6.\]

When \(x=-1.5\):

\[\frac{9}{4}a-\frac{3}{2}b+6=0 \quad\Rightarrow\quad 9a-6b=-24.\]

Multiplying \(4a+2b=-6\) by \(3\) gives \(12a+6b=-18\). Adding this to \(9a-6b=-24\):

\[21a=-42,\qquad a=-2.\]

Then:

\[4(-2)+2b=-6 \Rightarrow 2b=2 \Rightarrow b=1.\]

Hence,

\[a=-2,\qquad b=1,\qquad c=6,\]

and the relation is

\[\boxed{y=-2x^2+x+6}.\]

(c) The maximum is at the vertex. For \(y=ax^2+bx+c\), its \(x\)-coordinate is

\[x=\frac{-b}{2a}=\frac{-1}{2(-2)}=0.25.\]

\[y=-2(0.25)^2+0.25+6=6.125.\]

So the maximum value is

\[\boxed{6.125\text{ at }x=0.25}.\]

The graph may lead to approximate readings of \(x\approx0.2\) and \(y\approx6.2\), but \(0.25\) and \(6.125\) are the exact values from the equation.

(d) Set \(y=2\):

\[-2x^2+x+6=2.\]

\[2x^2-x-4=0.\]

\[x=\frac{1\pm\sqrt{(-1)^2-4(2)(-4)}}{2(2)} =\frac{1\pm\sqrt{33}}{4}.\]

\[\boxed{x\approx-1.2 \quad \text{or} \quad x\approx1.7}.\]

Examination reminder: Use the intercepts to find roots, the \(y\)-intercept to identify \(c\), and then substitute the known points into the quadratic to determine \(a\) and \(b\).