Question

How do you find the vertex and intercepts for `y=5x^2-30x+49`?

Answer 1

Vertex (3, 4)

Explanation:

x-coordinate of vertex:
`x = -b/(2a) = 30/10 = 3`
y-coordinate of vertex:
y(3) = 5(9) - 3(30) + 49 = 45 - 90 + 49 = 4
Vertex (3, 4).
Make x = 0, y-intercept = 49.
To find x-intercepts, make y = 0 and solve the quadratic equation:
`y = 5x^2 - 30x + 49 = 0.`
`D = b^2 - 4ac = 900 - 980 = -80.`
Since D < 0, there are no real roots (no x-intercepts). Since a > 0, the parabola opens upward and is completely above the x-axis.