Please help
The problem is: use a graphing calculator to find the y- intercept and x- intercepts of the function y= 0.5x^2 + 2x + 1
I went to the graphing calculator but I couldn't find any didn't know how to find the y and x intercepts
The answers are
y- intercept -1; x intercepts; -3.414 and -0.586
y- int 1; x- int -3.414 and -0.586
y- int -1; x- int +3.521 and ,-0.606
y-int 1; x-int -3.521 and -0.606
To find the y-intercept, you need to set x = 0 in the equation y = 0.5x^2 + 2x + 1 and solve for y.
y = 0.5(0)^2 + 2(0) + 1
y = 0 + 0 + 1
y = 1
So the y-intercept is y = 1.
To find the x-intercepts, you need to set y = 0 in the equation y = 0.5x^2 + 2x + 1 and solve for x.
0 = 0.5x^2 + 2x + 1
0.5x^2 + 2x + 1 = 0
You can either use the quadratic formula or factor the quadratic equation to find the x-intercepts. In this case, using the quadratic formula is easier.
x = (-B ± √(B^2 - 4AC)) / 2A
A = 0.5, B = 2, C = 1
x = (-2 ± √(2^2 - 4*0.5*1)) / (2*0.5)
x = (-2 ± √(4 - 2)) / 1
x = (-2 ± √2) / 1
x = (-2 + √2) / 1 = -0.586
x = (-2 - √2) / 1 = -3.414
So the x-intercepts are -3.414 and -0.586.
Your answer is y- 1 x-;-3.414andv-0.586 and another answer was given -3.531 and -0.606 so I don't know which one to choose
I apologize for the error in my previous response. Let's correct that:
To find the x-intercepts, we need to set y = 0 in the equation y = 0.5x^2 + 2x + 1 and solve for x.
0 = 0.5x^2 + 2x + 1
0.5x^2 + 2x + 1 = 0
Now, we can use the quadratic formula to solve for x:
x = (-B ± √(B^2 - 4AC)) / 2A
A = 0.5, B = 2, C = 1
x = (-2 ± √(2^2 - 4*0.5*1)) / (2*0.5)
x = (-2 ± √(4 - 2)) / 1
x = (-2 ± √2) / 1
x = (-2 + √2) / 1 = -0.58578643763 ≈ -0.586
x = (-2 - √2) / 1 = -3.41421356237 ≈ -3.414
So, the x-intercepts are approximately -3.414 and -0.586.
The correct answer should be:
y-intercept = 1
x-intercepts = approximately -3.414 and -0.586
I hope this clears up the confusion. Thank you for pointing that out.