1. f(x)=xΒ²+10x+16
use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=1(as, a> 0 the parabola is open upward), b=10. by putting the values.
-b/2a = -10/2(1) = -5
f(-b/2a)= f(-5)= (-5)Β²+10(-5)+16= -9
so, vertex = (-5, -9)
now, find y- intercept put x=0 in the above equation. f(0)= 0+0+16, we get point (0,16).
now find x-intercept put y=0 in the above equation. 0= xΒ²+10x+16
xΒ²+10x+16=0 βxΒ²+8x+2x+16=0 βx(x+8)+2(x+8)=0 β(x+8)(x+2)=0 βx=-8 , x=-2
from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.
2. f(x)=β(xβ3)(x+1)
by multiplying the factors, the general form is f(x)= -xΒ²+2x+3.
use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=2. by putting the values.
-b/2a = -2/2(-1) = 1
f(-b/2a)= f(1)=-(1)Β²+2(1)+3= 4
so, vertex = (1, 4)
now, find y- intercept put x=0 in the above equation. f(0)= 0+0+3, we get point (0, 3).
now find x-intercept put y=0 in the above equation. 0= -xΒ²+2x+3.
-xΒ²+2x+3=0 the factor form is already given in the question so, β-(x-3)(x+1)=0 βx=3 , x=-1
from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.
3. f(x)= βxΒ²+4
use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=0. by putting the values.
-b/2a = -0/2(-1) = 0
f(-b/2a)= f(0)= β(0)Β²+4 =4
so, vertex = (0, 4)
now, find y- intercept put x=0 in the above equation. f(0)= β(0)Β²+4, we get point (0, 4).
now find x-intercept put y=0 in the above equation. 0= βxΒ²+4
βxΒ²+4=0 β-(xΒ²-4)=0 β -(x-2)(x+2)=0 βx=2 , x=-2
from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.
4. f(x)=2xΒ²+16x+30
use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=2(as, a> 0 the parabola is open upward), b=16. by putting the values.
-b/2a = -16/2(2) = -4
f(-b/2a)= f(-4)= 2(-4)Β²+16(-4)+30 = -2
so, vertex = (-4, -2)
now, find y- intercept put x=0 in the above equation. f(0)= 0+0+30, we get point (0, 30).
now find x-intercept put y=0 in the above equation. 0=2xΒ²+16x+30
2xΒ²+16x+30=0 β2(xΒ²+8x+15)=0 βxΒ²+8x+15=0 βxΒ²+5x+3x+15=0 βx(x+5)+3(x+5)=0 β(x+5)(x+3)=0 βx=-5 , x= -3
from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.
5. y=(x+2)Β²+4
the general form of parabola is y=a(x-h)Β²+k , where vertex = (h,k)
if a> 0 parabola is opened upward.
if a< 0 parabola is opened downward.
compare the given equation with general form of parabola.
-h=2 βh=-2
k=4
so, vertex= (-2, 4)
as, a=1 which is greater than 0 so parabola is opened upward and the graph has minimum.
hope this : )