I really need help with these I have tried them and they are always wrong?

1. y-intercepts, where x = 0

y=x^2+2x-15

y = (x +5)(x – 3)

0 = (x +5)(x – 3)

0 = (x +5), 0 = (x – 3)

x = -5, x = 3

Write them: y-intercepts: (0,-5) & (0, 3)

1. y=x²+2x-15

y=0²+2(0)-15 <--plugged in 0 for x's y= -15 <--simplified 2. x²=-5+11 x²= 6 <--added like terms x²-6=0 <--moved 6 over by sub. 6 from both sides x= (-b +- sqrt(b²-4ac))/(2a) <--quadratic formula [+- = + or -] a(coefficient of x²)=1 b(coefficient of x)= 0 c(constant)= -6 x= (-0 +- sqrt(0²-4(1)(-6)))/(2*1) <--plugged in numbers x= (+- sqrt(24))/(2) <--simplified x= (+- 2sqrt(6))/(2) <--simplifed sqrt(24) x= (+- sqrt(6)) <--canceled out the 2's 3. h= -16t²+70t+4 h= -2(8t²-35t-2) <--factored out -2 t= (-35+- sqrt((-35)²-4(8)(-2)))/(2*-8) <--quadratic formula t= (-35+- sqrt(1225+64))/(-16) <--simplified t= (-35+- sqrt(1289))/(-16) <--simplified t= -0.06 or 4.43 seconds <--put into calculator t= 4.43 sec <--eliminated answer (can't have - seconds) 4. I can't actually graph this, but I'll show you how to do the table of values. x | y <--put 5 numbers on the x side of the table, preferably -2| -3 negative and positive numbers and zero. Plug in -1| 0 the numbers into the original equation to get the 0 | 1 y-column 1 | 0 2 | -3 Now just draw a graph and graph it. You might wanna check these over in case I made a mistake. Hope I helped!

Reese is incorrect. A quadratic function will only have ONE y-intercept.

1) Y intercept occurs when x=0. Plug in x=0

y intercept is -15

2) I think you have a typo, othersise you answer is +/- sqrt(6)

3) h=16t^2+70t +4 set h to zero, solve for t

h=(-70 +/- sqrt(70^2 + 4*16*4)) / 32.

4) Not enough information to complete this problem.

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