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During a theatre drama, a backdrop of building arches was used. The
nshape of the curve shown below can be represented by the polynomial
n$p(x) = -x^2 + 2x + 8$, where $x$ is the length (in feet) on stage level.
n(i) Determine the height of the arch.
n(ii) (a) Find zeroes of the polynomial $p(x)$. Which points on the
ngraph represent the zeroes?
nOR
n(ii) (b) Find the span of the arch on the stage floor.
n(iii) Write the coordinates of the point of intersection of the above curve
nwith the $y$-axis.
nshape of the curve shown below can be represented by the polynomial
n$p(x) = -x^2 + 2x + 8$, where $x$ is the length (in feet) on stage level.
n(i) Determine the height of the arch.
n(ii) (a) Find zeroes of the polynomial $p(x)$. Which points on the
ngraph represent the zeroes?
nOR
n(ii) (b) Find the span of the arch on the stage floor.
n(iii) Write the coordinates of the point of intersection of the above curve
nwith the $y$-axis.
Show SolutionHide Solution↓
(i) $p (1) = - (1)^2 + 2 \times 1 + 8 = 9$ (1 Mark)
So, height of the arch is $9$ feet.
(ii) (a) $p(x) = -x^2 + 2x + 8$
$= - (x-4)(x + 2)$ (1 Mark)
$\therefore$ zeroes are $-2$ and $4$. (1/2 Mark)
Points $B$ and $A$ on the graph represent the zeroes. (1/2 Mark)
OR
(ii) (b) $p(x) = -x^2 + 2x + 8$
$= - (x - 4)(x + 2)$ (1/2 Mark)
$\therefore$ zeroes are $-2$ and $4$. (1/2 Mark)
Hence, Coordinates of point $A$ and $B$ are $(4, 0)$ and $(-2, 0)$ respectively.
Span of the arch on the stage floor, $AB = 4 + 2 = 6$ (1 Mark)
So, span of the arch on the stage floor is $6$ feet.
(iii) $p (0) = - (0)^2 + 2 \times 0 + 8 = 8$ (1/2 Mark)
$\therefore$ the given curve intersects $y$-axis at $(0, 8)$ (1/2 Mark)
So, height of the arch is $9$ feet.
(ii) (a) $p(x) = -x^2 + 2x + 8$
$= - (x-4)(x + 2)$ (1 Mark)
$\therefore$ zeroes are $-2$ and $4$. (1/2 Mark)
Points $B$ and $A$ on the graph represent the zeroes. (1/2 Mark)
OR
(ii) (b) $p(x) = -x^2 + 2x + 8$
$= - (x - 4)(x + 2)$ (1/2 Mark)
$\therefore$ zeroes are $-2$ and $4$. (1/2 Mark)
Hence, Coordinates of point $A$ and $B$ are $(4, 0)$ and $(-2, 0)$ respectively.
Span of the arch on the stage floor, $AB = 4 + 2 = 6$ (1 Mark)
So, span of the arch on the stage floor is $6$ feet.
(iii) $p (0) = - (0)^2 + 2 \times 0 + 8 = 8$ (1/2 Mark)
$\therefore$ the given curve intersects $y$-axis at $(0, 8)$ (1/2 Mark)