Algebra Question

Question

Give the vertex, axis of symmetry, x-intercepts, y-intercept, domain, and range of the graph of the function:

f(x) = 3(x-5)² + 1

Answer 1

vertex is (5,1) the minimum point of the parabola. It's an upward opening parabola, with axis of symmetry as the vertical line running through the vertex, x=5. x intercepts are the roots, where y=0. where f(x)=0. set 3(x-5)^2 +1 equal to zero, and solve for x, that gives you the roots, zeroes, or x intercepts. 3(x^2-10x+25) +1 = 3x^2 - 30x +75 + 1 = 3x^2 -30x +76. Use the quadratic formula to solve for x, which gives you 0, 1 or 2 x intercepts. But we know the vertex, the minimum point is (5,1) so the parabola is above the x axis at all points,and never touches the x axis. There are no x-intercepts. The discriminate is negative, b^2-4ac<0 so the only roots are imaginary. It came close to the x axis though. Hand grenades and horse shoes, that gets a point or two, but not in math.

The y intercept(s) are found by setting x = 0 and solving for f(x) That gives f(x) =3(-5)^2 +1 = 3(25)+1=75+1=76 y=76 is the y intercept where the parabola intersects the y axis.

The original equation was in the general form of y=a(x-h)^2 + k, where (h,k) is the vertex. From there you can see most of the characteristics of the equation. a is relatively large, making the y intercept large, given the x intercept distance to the right of the origin.