1) gradient of line = Δ y ÷ Δ x
= (5 -2) ÷ (3 - (-6))
= ¹/₃

using the point-slope form (y-y₁) = m(x-x₁)
using (3,5)
(y - 5) = ¹/₃ (x -3)
   y - 5 = ¹/₃x - 1
⇒ <span> y = ¹/₃ x + 4 [OPTION D]

</span>2)    y = 2x + 5 .... (1)
<span>    </span>y = ¹/₂ x + 6 .... (2)

by substituting y in (1) for y in (2)

2x + 5 = ¹/₂ x + 6
   ³/₂ x = 1
x = ²/₃

by substituting found x (2)

    y = ¹/₂ (²/₃) + 6
y = ¹⁹/₃

∴ common point is (²/₃ , ¹⁹/₃) thus answer is FALSE [OPTION B]

3) Yes [OPTION A]
This is because the both have a gradient of 5 and if they have the same gradient then that means that the two lines are parallel to each other.

4) No [OPTION B]
Two lines are perpendicular if their gradients multiply to give - 1 and as such one is the negative reciprocal of the other. Since both gradients are ¹/₂ then they are actually parallel and not perpendicular.

6 0

2 years ago

Find the parabola with equation y = ax2 + bx whose tangent line at (3, 12) has equation y = 10x − 18. y = incorrect: your answer

pishuonlain [190]

For these questions to be true and the equation of the tangent to have an equal y to the equation of the parabola i guess there has to be a "c" and in that case integrate the equation of the tangent you will have a = 5 and b = -18 then you substitute in the equation of the parabola with the point you have you will find that "c" = 21 and so the equation of the parabola becomes y = 5x^2 - 18 x +21

3 0

2 years ago