Question

Given f(x) = 4 x^2 + 9 x , determine the points where
(a) the curve intercepts the x- axis;
(b) the tangent line to the curve is horizontal.

(a) The points where the curve intercepts the x- axis are . (The answers must be a POINTS on the curve; use parentheses for the points and commas to separate the points in the answer space.)
(b) The point where the tangent line to the curve is horizontal is . (The answer must be a POINT on the curve; separate the values with a comma and use parantheses in the answer space.)

Answer 1

a) This is a standard precalculus find the roots problem. Let f(x)=0 and solve.

b) Same as the problem we did a minute ago.

Answer 2

Make sure to show your answers in (x.y) form.

Answer 3

ok. it confused me because its worded differently

Answer 4

(1.125,15.1875) this is what i got, but its incorrect for the curve intercepts the x- axis;

Answer 5

this is not correct :(

Answer 6

hmm... did i did any calculation error?. the concept is there. weird..

Answer 7

wait, i mean(0,0) and ( -9/4,0)

Answer 8

(0,0),(-9/4,0)

Answer 9

intercept x-axis, f(x)=0
0=4x^2 + 9x
x=0, -9/4

when x=0, f(0)= 0
when x=-9/4, f(-9/4) =0

(0,0) and ( -9/4,0)

dy/dx= 8x+9
horizoltal, dy/dx=0
8x+9=0
x=-9/8.
when x=-9/8, f(-9/8)= -81/16

(-9/8, -81/16)

Answer 10

yes thats it, thanks a million

Answer 11

Remember, for a quadratic, you can find the vertex using the algebraic formula
(-b/2a, f(-b/2a)). It won't win you many points as a calculus method, but is an excellent check step. Again, it only works on quadratics.

Answer 12

thanks. :) everyone