Find the x-intercepts of the parabola given by the equation.

yapafw

yapafw

Answered question

2021-12-05

Find the x-intercepts of the parabola given by the equation. (If an answer does not exist, enter DNE.)
y=2x25x25
(x, y)=(smaller x-value)
(x, y)=(large x-value)
Find the minimum or maximum value of the quadratic function.
f(x)=5x22x

Answer & Explanation

Elizabeth Witte

Elizabeth Witte

Beginner2021-12-06Added 24 answers

Step 1
Given,
4=2x25x25
To find x-intercepts put, 4=0 in above equation
2x25x25=0
2x210x+5x25=0
2x(x5)+5(x5)=0(2x+5)(x+5)=0
2x+5=0, x5=0
x=52, x=5
Hence, x-intercept of parabola:
(x, y)=(52, 0) (smaller x-value)
(x, y)=(5, 0) (larger x-value)
Step 2
Given,
f(x)=5x22x
f(x)=10x22
To find critical point, f(x)=0
±0x2=0
x=210=15
Now, f(x)=10 y0 at x=15
So, the f(x) has a minimum at 15
Therefore, the minimum value is,
f(15)=5(15)225
f(15)=125×525=52525
f(15)=1525=15

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