To solve the given quadratic function problem, we will follow these steps:
(a) To express the quadratic function in standard form, we will complete the square.
(b) To find the vertex, we will use the vertex formula for a quadratic function in the form \( ax^2 + bx + c \), which is given by \( x = -\frac{b}{2a} \). We will then substitute this \( x \)-value back into the function to find the \( y \)-coordinate of the vertex.
For the \( x \)-intercepts, we will solve the equation \( f(x) = 0 \) using the quadratic formula. The \( y \)-intercept is found by evaluating \( f(0) \).
The given quadratic function is:
\[
f(x) = 2x^2 + x - 6
\]
The standard form of a quadratic function is:
\[
f(x) = a(x - h)^2 + k
\]
To convert the given function into standard form, we need to complete the square.
- Factor out the coefficient of \(x^2\) from the first two terms:
\[
f(x) = 2(x^2 + \frac{1}{2}x) - 6
\]
- Complete the square inside the parentheses. Take half of the coefficient of \(x\), square it, and add and subtract it inside the parentheses:
\[
\text{Coefficient of } x = \frac{1}{2}, \quad \left(\frac{1}{2} \times \frac{1}{2}\right)^2 = \frac{1}{16}
\]
- Add and subtract \(\frac{1}{16}\) inside the parentheses:
\[
f(x) = 2\left(x^2 + \frac{1}{2}x + \frac{1}{16} - \frac{1}{16}\right) - 6
\]
- Rewrite the expression as a perfect square trinomial:
\[
f(x) = 2\left(\left(x + \frac{1}{4}\right)^2 - \frac{1}{16}\right) - 6
\]
- Distribute the 2 and simplify:
\[
f(x) = 2\left(x + \frac{1}{4}\right)^2 - \frac{2}{16} - 6 = 2\left(x + \frac{1}{4}\right)^2 - \frac{1}{8} - 6
\]
- Combine the constants:
\[
f(x) = 2\left(x + \frac{1}{4}\right)^2 - \frac{49}{8}
\]
Thus, the standard form is:
\[
f(x) = 2\left(x + \frac{1}{4}\right)^2 - \frac{49}{8}
\]
The vertex form of a quadratic function is:
\[
f(x) = a(x - h)^2 + k
\]
From the standard form we derived:
\[
f(x) = 2\left(x + \frac{1}{4}\right)^2 - \frac{49}{8}
\]
The vertex \((h, k)\) is:
\[
h = -\frac{1}{4}, \quad k = -\frac{49}{8}
\]
Thus, the vertex is:
\[
(x, y) = \left(-\frac{1}{4}, -\frac{49}{8}\right)
\]
To find the \(x\)-intercepts, set \(f(x) = 0\):
\[
2x^2 + x - 6 = 0
\]
Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 2\), \(b = 1\), and \(c = -6\).
Calculate the discriminant:
\[
b^2 - 4ac = 1^2 - 4 \times 2 \times (-6) = 1 + 48 = 49
\]
Calculate the roots:
\[
x = \frac{-1 \pm \sqrt{49}}{4} = \frac{-1 \pm 7}{4}
\]
The roots are:
\[
x_1 = \frac{-1 + 7}{4} = \frac{6}{4} = \frac{3}{2}
\]
\[
x_2 = \frac{-1 - 7}{4} = \frac{-8}{4} = -2
\]
Thus, the \(x\)-intercepts are:
\[
(x, y) = (-2, 0) \quad \text{and} \quad (x, y) = \left(\frac{3}{2}, 0\right)
\]
To find the \(y\)-intercept, set \(x = 0\):
\[
f(0) = 2(0)^2 + 0 - 6 = -6
\]
Thus, the \(y\)-intercept is:
\[
(x, y) = (0, -6)
\]
- Standard form: \(\boxed{f(x) = 2\left(x + \frac{1}{4}\right)^2 - \frac{49}{8}}\)
- Vertex: \(\boxed{(x, y) = \left(-\frac{1}{4}, -\frac{49}{8}\right)}\)
- \(x\)-intercepts: \(\boxed{(x, y) = (-2, 0)}\) and \(\boxed{(x, y) = \left(\frac{3}{2}, 0\right)}\)
- \(y\)-intercept: \(\boxed{(x, y) = (0, -6)}\)
![A quadratic function (f) is given.
[f(x)=2 x^2+x-6]
(a) Express (f) in standard form.
[f(x)=]
(b) Find its vertex and (x)-and (y)-intercept(s) of (f). (If an answer does not exist, enter DNE.)
vertex
[(x, y)=()]
(x)-intercepts ((x, y)=) ) (smaller (x)-value)
((x, y)=()) (larger (x)-value)
(y)-intercept ((x, y)=) )](https://thumbnail.justsolvely.com/seoQuestionThumb/d8706e1b-3ea3-406f-a040-48278a9bd6c7.webp)
Answered
