Above is the graph of the quadratic function \(y = ax^{2} + bx + c\) where a, b and c are constants. Using the graph, find : (a)(i) the scales on both axes ...

Answer Details

There is an inconsistency in the supplied reference answer. The coordinates \(D(0,1)\), \(E(1,-2)\), and \(G(3,4)\) give the quadratic \(y=2x^2-5x+1\). This equation has its vertex at \(x=1.25\), not at \(x=1\). Also, for \(-3\leq x\leq 5\), its greatest value is \(34\), not \(33.5\).

Graph consistent with the given coordinates and equation

(a)(i) Scales

• On the \(x\)-axis, \(2\text{ cm}\) represents \(1\) unit.
• On the \(y\)-axis, \(2\text{ cm}\) represents \(5\) units.

(a)(ii) Line of symmetry

The line of symmetry passes through the vertex. From the graph, the minimum is halfway between \(x=1\) and \(x=1.5\), so

\[x=1.25\]

Do not assume that \(E(1,-2)\) is the vertex. It is a point on the curve, but the curve continues slightly lower before turning upwards.

(a)(iii) Roots

The roots are the \(x\)-coordinates where the curve crosses the \(x\)-axis. Reading from the graph gives approximately

\[x\approx0.25\quad\text{and}\quad x\approx2.25.\]

Using the equation found in part (b), the more accurate values are

\[x=\frac{5-\sqrt{17}}{4}\approx0.219,\qquad x=\frac{5+\sqrt{17}}{4}\approx2.281.\]

(b) Finding \(a\), \(b\), and \(c\)

Use \(y=ax^2+bx+c\) and substitute the three stated points.

Using \(D(0,1)\):

\[1=a(0)^2+b(0)+c\]

\[c=1\]

Using \(E(1,-2)\):

\[-2=a+b+1\]

\[a+b=-3\qquad\text{(1)}\]

Using \(G(3,4)\):

\[4=9a+3b+1\]

\[9a+3b=3\]

\[3a+b=1\qquad\text{(2)}\]

Subtract equation (1) from equation (2):

\[(3a+b)-(a+b)=1-(-3)\]

\[2a=4\]

\[a=2\]

Then substitute \(a=2\) into \(a+b=-3\):

\[2+b=-3\]

\[b=-5\]

Therefore,

\[\boxed{y=2x^2-5x+1}\]

(c) Greatest value of \(y\) for \(-3\leq x\leq5\)

Since \(a=2>0\), the parabola opens upwards. Its vertex is a minimum, so the greatest value in the given closed interval must be at an endpoint.

At \(x=-3\):

\[y=2(-3)^2-5(-3)+1=18+15+1=34\]

At \(x=5\):

\[y=2(5)^2-5(5)+1=50-25+1=26\]

Since \(34>26\), the greatest value is

\[\boxed{34}\]

Examination reminder: For a quadratic opening upwards, check both endpoints when asked for the greatest value over a restricted interval. The vertex gives the minimum, not the maximum.