find teh x and y intercepts

Question

f(x)= 3x^2+12x+1

a ((-6+-squareroot39)/3,0),(0,-1)

b ((-12+- squareroot39)/3),0), (0,1)

c ((-6+- squareroot 33)/6,0),(0,-1)

d ((-6+- squareroot 33)/),0), (0,1)

Answer 1

When x=0, y=0+0+1=1 when this value of x is put into the function. So the intercept of the vertical axis is (0,1). The solution of the quadratic, when f(x)=0, is found by applying the formula, i.e., x=(-12+/-sqrt(144-12))/6=(-12+/-sqrt(132))/6. 132=4*33, so this becomes (-12+/-2*sqrt(33))/6. Dividing top and bottom by 2 we get (-6+-sqrt(33))/3. This shows the two points where the curve intercepts the x axis, where f(x)=0. So d is the correct option, although there's a missing 3. The correct answer is ((-6+/-sqrt(33))/3,0), (0,1), where plus-or-minus is shown by +/-.

Answer 2

3x^2 +12x +1=0

quadratik equashun giv roots=-0.085145784 & -3.914854216

-b/2a=-2, b^2-41c=132, sqrt(132)=11.489125293, sqrt/2a=1.9148542155