let y = 9x^2 + 24x + 16
for the y-intercept , let x = 0 , --->y = 16
for the x-intercept, let y = 0
9x^2 + 24x + 16 = 0
you could use the quadratic formula, but I noticed the above is a perfect square, so
(3x+4)^2 = 0
3x+4 = 0
x = -4/3
So there is one x-intercept at x = -4/3
(the parabola has its vertex at (-4/3, 0) and crosses the y-axis at (0,16) )
find the x and y intercepts
f(x)=9x^2+24x+16
Help
2 answers
This crosses the y axis when x=0 so
y=16
This crosses the x axis when y=0
0=9x^2+24x+16
0=(x+4/3)^2
so x=-4/3
y=16
This crosses the x axis when y=0
0=9x^2+24x+16
0=(x+4/3)^2
so x=-4/3