How do you find the vertex and intercepts for #f(x)=12.25x^2 - 52.5x +110.25#?

Question

Isaiah Anthony · Accepted Answer

Vertex (2.14, 54)

#f(x) = 12.25x^2 - 52.5x + 110.25# x-coordinate of vertex: #x = -b/(2a) = 52.5/24.5 = 2.14# y-coordinate of vertex: #f(2.14) = 12.25(2.14)^2 - 52.5(2.14) + 110.25 =# #= 56.1 - 112.35 + 110.25 = 54# Vertex (2.14, 54) To find the 2 intercepts, solve the quadratic equation: #f(x) = 12.25x^2 - 52.5x + 110.25 = 0# #D = d^2 = b^2 - 4ac = 2756.25 - 5402.25 = - 2646 < 0# Since D < 0, there are no x-intercepts. The parabola graph of f(x) stays completely above the x-axis.

Julia Adams · Answer

undefined To find the vertex and intercepts for the quadratic function $f(x) = 12.25x^2 - 52.5x + 110.25$, follow these steps:

1. **Vertex**: Use the formula $x = \frac{-b}{2a}$ to find the x-coordinate of the vertex. Then, substitute this x-value into the function to find the corresponding y-coordinate.

2. **Intercepts**:
   - **x-intercepts**: Set $f(x) = 0$ and solve for $x$.
   - **y-intercept**: Set $x = 0$ and find $f(0)$.

Let's proceed with the calculations:

1. **Vertex**:
   $a = 12.25$, $b = -52.5$.
   Substitute into the formula: $x = \frac{-(-52.5)}{2 \times 12.25} = \frac{52.5}{24.5}$.
   Calculate: $x = 2.14$.
   Substitute $x = 2.14$ into the function: $f(2.14) = 12.25(2.14)^2 - 52.5(2.14) + 110.25$.
   Calculate: $f(2.14) ≈ 101.36$.
   So, the vertex is approximately $ (2.14, 101.36)$.

2. **Intercepts**:
   - **x-intercepts**:
     Set $f(x) = 0$: $12.25x^2 - 52.5x + 110.25 = 0$.
     Solve the quadratic equation to find the roots.
   - **y-intercept**:
     Set $x = 0$: $f(0) = 12.25(0)^2 - 52.5(0) + 110.25$.
     Calculate: $f(0) = 110.25$.

Therefore, the vertex is approximately $ (2.14, 101.36)$, and the y-intercept is $ (0, 110.25)$. To find the x-intercepts, solve the quadratic equation.