Find a formula for the quadratic function whose graph has a y-intercept of y=7 and x-intercept at x=1,6. f(x)

Answer 1

A quadratic function is a second-degree polynomial function that looks like f(x) = ax² + bx + c

Y-intercept means the function crosses the Y-axis at that point so the X value will be zero at that point, so if we replace x with zero in our equation and y or f(x) with 7, our c value in fact as the Y-intercept will be 7: c=7

X-intercept means the function crosses X-axis at that/those points and thus y or f(x) value will be zero at that/those points. Thus, we get two forms of equations if we put those two X-intercepts (x=1 and x=6) in our formula:

x = 1 ⇒ 0 = a + b + 7 ⇒ -b-7 = a

x = 6 ⇒ 0 = 36a + 6b + 7 ⇒ 0 = 36(-b-7) + 6b + 7 ⇒ 0 = -36b-252+6b+7 ⇒ 0 = -30b-245 ⇒ -30b = 245 ⇒ b = -245/30

-b-7 = a ⇒ -(-245/30) -7 = a ⇒ a = (245/30) - 7 ⇒ a = 35/30

Finally, our quadratic formula will be:

f(x) = (35/30)x² + (-245/30)x + 7

f(x) = (35/30)x² - (245/30)x + 7

Answer 2

I assume you mean TWO x intercepts---at x = 1 and 6

If it's a quadratic, it's a parabola. If there are two x intercepts, then it opens either up or down.

Let's us the standard form y + b = k*(x + a)² With x intercepts at 1 and 6, the axis is halfway in between---at x = 3.5.

So, now we have y + b = k*(x - 3.5)²

We have 2 unknowns: b and k. The trick is to use the given values of the intercepts.

First, if y = 0, x is either 1 or 6. This gives us:

b = k*(±2.5)²

b = 6.25k

Finally, use the y intercept to get another equation:

7 - b = k(3.5)²

Replace b using the previous equation:

7 - 6.25k = 12.25k

6k = 7

k = 7/6

and then b = 6.25*7/6 = 7.2917 (more precisely 25*7/24)

The equation is then y + 7.2917 = 7/6 * (x - 3.5)²

CHECK by graphing