Find the x-intercepts of the parabola with vertex (4,-1) and y-intercept (0,15). .

Question

Find the x-intercepts of the parabola with vertex (4,-1) and y-intercept (0,15).
Write your answer in this form: (x1,y1),(x2,y2).
If necessary, round to the nearest hundredth.

Answer 1

We have to use the general formula for a parabolic function: f(x) = a(x-h)²+k, which will give you the vertex in the form (h, k).  Since we know the vertex is at (4,-1) we now have a function: f(x) = a(x-4)²-1. Next, since we know the parabola passes through point (0, 15), we can plug these into our function so solve for a --> 15 = a(0-4)²-1...therefore, a must equal 1.  Now we have the function f(x) = (x-4)²-1 and we can use this to solve for the x intercepts by letting f(x) = 0 and solving for x.  We should end up with x-intercepts of 3 and 5.

Answer 2

We can modify express a generic quadratic function in the following way:

 

f(x) = A(x-v)^2+C, where x=v is the x coordinate of the vertex (in this case v=4)...and plug in the known points to find A and C.

 

f(4)=-1=A(4-4)^2+C, which means C=-1

f(0)=15=A(0-4)^2-1, which means A=1

 

Thus f(x)=(x-4)^2 - 1= x^2-8x+15

 

Now we can apply the quadratic formula to find the roots of x=intercepts of this quadratic function (parabola).

 

Hope this helps!