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Write an equation for a parabolic with x-intercepts at (1/4,0) which passes through the point (0,5)

User Jan Paepke
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Final answer:

The equation of the parabola is y = -84x^2 + x + 5.

Step-by-step explanation:

To find the equation of a parabola with x-intercepts at (1/4,0) which passes through the point (0,5), we can use the standard form of a parabola equation, y = ax^2 + bx + c. Since the x-intercepts are (1/4,0), we can write two equations:

1) When x = 1/4, y = 0 :

0 = (1/4)^2a + (1/4)b + c

2) When x = 0, y = 5 :

5 = 0^2a + 0b + c

From the second equation, we can see that c = 5. Substituting this value into the first equation, we can solve for a and b.

0 = (1/4)^2a + (1/4)b + 5

0 = (1/16)a + (1/4)b + 5

Multiplying both sides of the equation by 16 to eliminate the fractions:

0 = a + 4b + 80

From this equation, we can see that a = -4b - 80.

Substituting this expression for a in the second equation:

5 = 0^2(-4b - 80) + 0b + 5

5 = 0 + 0 + 5

So, b can be any real number. Let's take b = 1. Substituting this value for b in the expression for a:

a = -4(1) - 80 = -4 - 80 = -84

So, the equation of the parabola is y = -84x^2 + x + 5.

User Neurix
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