Write the equation for the parabola that has x− intercepts (−4,0) and (1.5,0) and y− intercept (0,−15).

Answer 1

Since the x-intercepts occur at the points (-4,0) and (1.5,0), the quadratic polynomial f(x) representing this parabola has factors (x+4) and (x-1.5) = (x-3/2). Hence f(x) has the form

f(x) = a(x+4)(x-3/2), where a is nonzero number. Since the y-intercept is (0,-15), we can solve for a by substituting 0 for x and -15 for f(x):

-15 = a(0+4)(0-3/2)

15 = -6a

(-15)/(-6) = a

5/2 = a.

Thus we can write the equation for the parabola as

y = (5/2)(x+4)(x-3/2), or, after multiplying,

y= (5/2)x²+(25/4)x-15. This is an acceptable answer. We show how to write the equation in vertex form below.

The equation above is written in the form

y = ax^2 + bx + c. (a = 5/2, b = 25/4, c = -15).

If the desired form for the equation of the parabola is the vertex form

y = a(x-h)²+k, where a is the same a as above, and (h,k) is the vertex of the parabola,

we can use the formulas h = -b/(2a) and k = c - b²/(4a) to find the coordinates of the vertex.

In this case, h = -(25/4)/(2(5/2)) = -5/4 and k = -15 - (25/4)²/(4(5/2) = -605/32.

Therefore, in vertex form, we can write the equation as

y = (5/2)(x+5/4)²-605/32.