Given: y = (x + 3)^2 + 5y=(x+3)2+5
The vertex can be found using the vertex form: y = (x - h)^2 + ky=(x−h)2+k,
where the vertex is (h, k)(h,k)
In the given equation the color(blue)("vertex": (-3, 5))vertex:(−3,5)
The yy-intercept is found setting x = 0x=0:
y = (0 + 3)^2 + 5 = 14y=(0+3)2+5=14
color(blue)(y"-intercept "(0, 14))y-intercept (0,14)
Find xx-intercepts by setting y = 0y=0 and factor or use the quadratic formula.
Put the equation in Ax^2 + Bx + C = 0Ax2+Bx+C=0 form. Use the squaring formula (a + b)^2 = a^2 + 2ab + b^2(a+b)2=a2+2ab+b2:
y = y=x^2 + 6x + 9 + 5#
y = x^2 + 6x + 14y=x2+6x+14
Using the quadratic formula x = (-B +- sqrt(B^2 -4AC))/(2A)x=−B±√B2−4AC2A:
x = (-6 +- sqrt(36 -4(1)(14)))/(2) = -3 +- (sqrt(-24))/2x=−6±√36−4(1)(14)2=−3±√−242
Since there is a negative under the radical, the xx values are imaginary. This means there are color(blue)("no "x"-intercepts")no x-intercepts.
To plot more points, you would need to do point plotting. Since xx is the independent variable, you would select different xx values and calculate the yy values.
ul(" "x " "| " "y" ")
" "-6" "|" "14
" "-4" "|" "6
" "-2" "|" "6
" "-1" "|" "9
graph{(x + 3)^2 + 5 [-20, 10, -4, 20]}
