Question

please solve this question​

Answer 1

(a)

(i) for the parabola y = x² + 1

=> x² = y - 1

=> (x - 0)² = 4$(\frac{1}{4})$ (y - 1)

on comparing with the vertex form

(x - h)² = 4a(y - k)

(h, k) = (0, 1)

point C is the vertex

Hence, coordinates of point C are (0, 1).

(ii) for point of intersection of line y = 4 - x with x - axis, y = 0

hence, x = 4

for point of intersection on y - axis, x = 0

hence, y = 4

Hence, the line y = 4 - x intersects the x and y axes at x = 4 and y = 4 respectively.

(b)

we know that, for any line ax + by + c = 0

gradient = $\frac{-a}{b}$

hence, gradient of line y = 4 - x

i.e.

x + y - 4 = 0

is

$\frac{-1}{1} = - 1$

Hence, gradient of the line y = 4 - x is -1.

(c)

for gradient of the graph y = x² + 1 to be negative;

we know that, gradient of a curve at a point = $\frac{dy}{dx}$

hence,

if

$\frac{dy}{dx} < 0 \\ \frac{d}{dx} ( {x}^{2} + 1) < 0 \\ 2x + 0 < 0 \\ 2x < 0 \\ x < 0$

Hence, range of x for which gradient of the graph y = x² + 1 is negative is (-∞, 0)

(d)

to find their intersection;

solve the equations simultaneously

y = x² + 1

y = 4 - x

x² + 1 = 4 - x

x² + x - 3 = 0

this hence gives the abscissae of intersection of the straight line and the curve.

(e)

x² + x - 3 = 0

x² + x = 3

x² + 2$\frac{1}{2}x$ + ¼ = 3 + ¼

(adding ¼ to both sides)

(x + ½)² = $\frac{13}{4}$

x + ½ = $\sqrt(\frac{13}{4})$

x + ½ = $± \frac{\sqrt{13}}{2}$

x + 0.5 = $± \frac{3.6055}{2}$

x + 0.5 = ± 1.80275

x = ± 1.803 - 0.5

x = 1.803 - 0.5 or x = -1.803 - 0.5

x ≈ 1.303 or x ≈ -2.303

Hence, solutions of x² + x - 3 = 0 are 1.303 and -2.303.

(f)

from (e) abscissae of A and B are -2.303 and 1.303 respectively

substituting in y = 4 - x

y(B) = 4 - 1.303 or y(A) = 4 - (-2.303)

y(B) = 2.697 or y(A) = 6.303

now, A(x, y) = (-2.303, 6.303)

and B(x, y) = (1.303, 2.697)

hence, coordinates of midpoint of segment AB

= $( \frac{x(A) + x(B)}{2} , \: \frac{y(A) + y(B)}{2})$

= $(\frac{-2.303 + 1.303}{2}, \: \frac{6.303 + 2.697}{2})$

= $( \frac{-1}{2}, \: \frac{9}{2} )$

= (-0.5, 4.5)

Hence, coordinates of the midpoint of the segment AB are (-0.5, 4.5).

p.s. please tell me where is this question from..