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

Question

Ethan Adkins · Accepted Answer

Vertex (3, 4)

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.

Julia Adams · Answer

undefined To find the vertex and intercepts for the quadratic function $y = 5x^2 - 30x + 49$, follow these steps:

1. **Vertex:** The vertex of a quadratic function in the form $y = ax^2 + bx + c$ is given by the formula $(-b/2a, f(-b/2a))$, where $f(x)$ is the function. For the given function $y = 5x^2 - 30x + 49$, $a = 5$ and $b = -30$. The x-coordinate of the vertex is $-(-30) / (2 \cdot 5) = 3$. Substituting $x = 3$ into the function gives $y = 5(3)^2 - 30(3) + 49 = 4$. Therefore, the vertex is at the point $(3, 4)$.

2. **Intercepts:**
   - **x-intercepts:** To find the x-intercepts, set $y = 0$ and solve for $x$. 
     $0 = 5x^2 - 30x + 49$
     This quadratic equation does not factor easily, so you can use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the roots.
     $x = \frac{30 \pm \sqrt{(-30)^2 - 4 \cdot 5 \cdot 49}}{2 \cdot 5}$
     $x = \frac{30 \pm \sqrt{900 - 980}}{10}$
     $x = \frac{30 \pm \sqrt{-80}}{10}$
     Since the square root of a negative number is not real, the function does not have x-intercepts.
   
   - **y-intercept:** To find the y-intercept, set $x = 0$ and solve for $y$.
     $y = 5(0)^2 - 30(0) + 49$
     $y = 49$
     Therefore, the y-intercept is at the point $(0, 49)$.

In summary, the vertex of the function is at $(3, 4)$, and it does not have x-intercepts but has a y-intercept at $(0, 49)$.