A quadratic curve intersects the axes at (–3, 0), (3, 0) and (0, 18). Work out the equation of the curve

Using the equation y = ax2 + bx + cCreate 3 separate equations:-a(3)2 + b(3) + 18 = 0 -a(-3)2 + b(-3) + 18 = 0
-9a+3b = -18-9a - 3b = -18
add the equations:-9a-9a-3b+3b=-18-18-18a = -36a = 2
Substitute a into equation to find b:
-2(9) - 3b = -18b =0
Therefore,
y = -2x2 + 18

Answered by Stanley S. Maths tutor

6198 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Expand 5yx^2 (3x^3 - 5xy + wy^2)


The point P has coordinates (3,4), Q has the coordinates (a,b), a line perpendicular to PQ is given by the equation 3x+2y=7. Find an expression for b in terms of a


Solve x^2 -4x=12


Find the gradient of the line on which the points (1,3) and (3,4) lie and find the y-coordinate of the line at x = 7.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy