Question

find the vertex,all intercepts and graph y=5-12x-3x^2

Answer 1

y=5-12x-3x^2 is a parabola. The roots when y=0 give the intercepts on the x axis. The equation 3x^2+12x-5=0 has roots according to the quadratic formula: (-12+/-sqrt(144+60))/6 ((-b+/-sqrt(b^2-4ac))/2a is the standard formula). The two irrational roots are 0.38048 and -4.38048. The intercept on the y axis is 5 (at x=0). The vertex is -b/2a, so that gives us x=-12/6 = -2 and y=17. The graph resembles an inverted U and is above the x axis for x between the two roots and it is symmetrical about the line x=-2 (the vertex), midway between the two roots.This description should help you to plot the graph.

Answer 2

y=-3x^2-12x+5

step 1...find zeroes...yuze quadratik equashun

zeroes=-4.380476143 & +0.380476143

-b/21=-2, b^2-4ac=204, sqrt(204)=14.282856857, sqrt/21=-2.38047614284

step 2...tu find top or bottom...dy/dx=-6x-12, slope=0 at 6x=-12, or x=-2

at x=-2, y=-3*4 -2*(-12)+5...=-12+24+5=12+5=17, so top=(-2,17)

y-intersept...x=0, so y=5 or(0,5)

x-intersept...y=0, so -3x^2-12x+5=0...thats the 2 zeroes up abuv