Hello :
the tangent has same slope of the line 16x -y +5 = 0 because parallel.
y = 16x + 5
the slope is : 16
but : f'(x) = 16
f'(x) = 16x^3 - 16x +16
solve : 16x^3 - 16x +16 = 16
x^3 - x +1 =1
x^3 - x = 0
x(x² - 1 ) =0
x(x -1 )(x +1) =0
x=0 or x-1 =0 or x +1= 0
x=0 or x = 1 or x = -1
if : x = 0 : y = 4(0)^4 -8(0)^2+16(0) +7 =7
if : x =1 : y = 4(1)^4 -8(1)^2+16(1) +7 = 19
if : x = -1 : y = 4(-1)^4 -8(-1)^2+16(-1) +7 = -13
<span>the points : A(0,7) , B(1 , 19) , c ( -1 , -13)</span>
The answer would be m^2+3m+2=0 because the first “part” of the equation (x^2-5) is equal to m; in the equation it’s squared so it can be replaced by m^2. next, 3m is the same as 3(x^2+5) which is what you have when you see 3x^2+15. finally, you have to bring the -2 over from the other side so you have positive 2. hope this helps!
