Thank you for answering my question.
Can you steer me to making a quadratic from
1/4(x+3)^2-4
Then I can get the intercepts.
Thanks
Just multiply it out to put it in the standard form,
ax^2 + b^2 + c
(1/4)(x+3)^2-4 = (1/4)(x^2 + 6x + 9) -4
= (1/4)x^2 + (3/2)x - 7/4
You're a star!
Thanks
You're welcome! I'm glad I could steer you in the right direction. To find the intercepts of a quadratic equation, you need to set the equation equal to zero and solve for x. In this case, the quadratic equation is:
(1/4)x^2 + (3/2)x - 7/4 = 0
To simplify the equation, you can multiply through by 4 to get rid of the fraction:
x^2 + 6x - 7 = 0
Now, you can use the quadratic formula to find the values of x that make the equation equal to zero. The quadratic formula is:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this equation, a = 1, b = 6, and c = -7. Plug these values into the quadratic formula and solve for x. You will get two possible values for x, which correspond to the x-intercepts of the quadratic equation.