Could someone verify if I got two answers to two problems right?
f(x)=x^2-9x+15 find x and y intercepts
-9+square root of 21/2,0
-9-square root of 21/2,0
x^2-x+3=0 solve and find x-intercepts
(1+square root of -11/2,0)
(-1-square root of -11/2,0)
for ax^2 + bx + c = 0,
x = (-b ± √(b^2 - 4ac) )/(2a)
so for yours, for the x-intercepts
x = (+9 ± √(81-4(1)(15))/2
= (9 ± √21)/2
for the y-intercept, let x=0
f(0) = 0-9+15 = 15
for the second:
x-intercept: (1 + √-11)/2 , (you had 1 + √-11/2 ---> only the last term was divided by 2 )
y-intercept : 3
Hi
I don't understand how you are getting the y intercept. I am so confused about which numbers are solutions, vertex, x-intercepts, y-intercepts etc.
Can you write out an example of an equation and how you get each number from a formula. I have been in this class for nine weeks and with all the information I have at my disposal I still am confused.
To verify if your answers are correct, let's first find the x and y-intercepts for the function f(x) = x^2 - 9x + 15.
To find the x-intercepts, we need to set f(x) = 0 and solve for x. So, we have:
x^2 - 9x + 15 = 0
To factor this quadratic equation, we look for two numbers whose sum is -9 and whose product is 15. In this case, the numbers are -3 and -5. Factoring, we get:
(x - 3)(x - 5) = 0
Setting each factor to zero, we obtain two possible solutions:
x - 3 = 0 => x = 3
x - 5 = 0 => x = 5
So, the x-intercepts are x = 3 and x = 5.
To find the y-intercept, we substitute x = 0 into the equation f(x) = x^2 - 9x + 15:
f(0) = 0^2 - 9(0) + 15
f(0) = 0 + 0 + 15
f(0) = 15
Therefore, the y-intercept is (0, 15).
Based on these calculations, your answers for the x and y-intercepts of f(x) = x^2 - 9x + 15 are correct:
-9 + √21/2, 0
-9 - √21/2, 0
Now let's verify the solutions for the second problem, which involves solving the equation x^2 - x + 3 = 0 and finding the x-intercepts.
Unfortunately, the solutions you provided: (1 + √(-11)/2, 0) and (-1 - √(-11)/2, 0) are not valid in the real number system because √(-11) results in an imaginary number. Therefore, there are no real solutions, meaning this quadratic equation has no x-intercepts in the real number system.
However, if you were working within the complex number system, the solutions would be valid.