A quadratic function has \( x \)-intercepts at \( \left(-\frac{1}{2}, 0\right) \) and \( (-1,0) \). The point, \( (4,90) \), lies on the parabola. Complete the statements. The value of \( a \) is 14 The equation of the quadratic function is \( f(x)= \) \( \begin{array}{cll}2 & (2 x+1)(x+2) & \times \\ \text { in factored form. }\end{array} \)

To find the equation of the quadratic function given the \( x \)-intercepts and a point on the parabola, we can follow these steps: 1. **Identify the roots**: The \( x \)-intercepts are given as \( -\frac{1}{2} \) and \( -1 \). This means the quadratic can be expressed in factored form as: \[ f(x) = a(x + \frac{1}{2})(x + 1) \] where \( a \) is a constant that will be determined later. 2. **Expand the factored form**: We can expand the expression: \[ f(x) = a\left(x + \frac{1}{2}\right)(x + 1) = a\left(x^2 + x + \frac{1}{2}x + \frac{1}{2}\right) = a\left(x^2 + \frac{3}{2}x + \frac{1}{2}\right) \] 3. **Use the point (4, 90)**: We know that the point \( (4, 90) \) lies on the parabola, which means: \[ f(4) = 90 \] Substituting \( x = 4 \) into the equation gives: \[ f(4) = a\left(4^2 + \frac{3}{2}(4) + \frac{1}{2}\right) = 90 \] Simplifying this: \[ f(4) = a\left(16 + 6 + \frac{1}{2}\right) = a\left(22.5\right) = 90 \] Now, we can solve for \( a \): \[ a \cdot 22.5 = 90 \implies a = \frac{90}{22.5} = 4 \] 4. **Write the final equation**: Now that we have \( a = 4 \), we can write the equation of the quadratic function: \[ f(x) = 4\left(x + \frac{1}{2}\right)(x + 1) \] 5. **Convert to standard form**: To express it in standard form, we can expand it: \[ f(x) = 4\left(x^2 + \frac{3}{2}x + \frac{1}{2}\right) = 4x^2 + 6x + 2 \] Thus, the completed statements are: - The value of \( a \) is \( 4 \). - The equation of the quadratic function is \( f(x) = 4\left(x + \frac{1}{2}\right)(x + 1) \) in factored form.