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 relation illustrated is a downward parabola. Reading from the graph, the curve cuts the y-axis at \(y = 6\), cuts the x-axis at \(x = -1.5\) and \(x = 2\), and turns over near \(x = 0.25\). The annotated graph below shows these key features, which we read off to answer each part.

(a) Roots of \(ax^{2}+bx+c=0\)

The roots are where the curve crosses the x-axis (\(y=0\)). From the graph these are:

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

(b) Values of a, b and c, and the relation

The curve cuts the positive y-axis at \(6\), so the intercept gives

\[ c = 6. \]

Substitute the coordinates of the two x-axis crossings, \(Q(2,\,0)\) and \(P(-1.5,\,0)\), into \(y = ax^{2}+bx+c\) with \(y = 0\) and \(c = 6\).

At \(x = 2\):

\[ a(2)^{2} + b(2) + 6 = 0 \implies 4a + 2b = -6 \quad \text{...(1)} \]

At \(x = -1.5\):

\[ a(-1.5)^{2} + b(-1.5) + 6 = 0 \implies \tfrac{9a}{4} - \tfrac{3b}{2} = -6 \implies 9a - 6b = -24 \quad \text{...(2)} \]

Multiply (1) by 3:

\[ 12a + 6b = -18 \quad \text{...(1a)} \]

Add (1a) and (2):

\[ 21a = -42 \implies a = -2. \]

Substitute \(a = -2\) into (1):

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

Therefore \((a,\,b,\,c) = (-2,\,1,\,6)\), and the relation illustrated by the graph is

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

(c) Maximum value of y and the corresponding x

The maximum occurs at the turning point. Completing the derivative (or reading the vertex from the graph):

\[ \frac{dy}{dx} = -4x + 1 = 0 \implies x = \tfrac{1}{4} = 0.25. \]

Substituting \(x = 0.25\):

\[ y_{\max} = -2(0.25)^{2} + 0.25 + 6 = -0.125 + 0.25 + 6 = 6.125. \]

So the maximum value is \(y_{\max} \approx 6.13\) (about \(6.1\)), occurring at \(x = 0.25\), which agrees with the peak read from the graph.

(d) Values of x when \(y = 2\)

Draw the horizontal line \(y = 2\) and read where it meets the curve, or solve:

\[ -2x^{2} + x + 6 = 2 \implies -2x^{2} + x + 4 = 0 \implies 2x^{2} - x - 4 = 0. \]

Using the quadratic formula:

\[ x = \frac{-(-1) \pm \sqrt{(-1)^{2} - 4(2)(-4)}}{2(2)} = \frac{1 \pm \sqrt{33}}{4} = \frac{1 \pm 5.745}{4}. \] \[ x = \frac{1 + 5.745}{4} = 1.69 \quad \text{or} \quad x = \frac{1 - 5.745}{4} = -1.19. \]

So when \(y = 2\), \(x \approx -1.2\) or \(x \approx 1.7\), matching the two points where the line \(y = 2\) cuts the curve on the graph.