To find the [tex]\(x\)[/tex]-intercepts and the vertex of the quadratic function [tex]\(f(x) = 1.69x^2 + 11.7x + 20.25\)[/tex], let's follow the steps thoroughly.
### Finding the [tex]\(x\)[/tex]-intercepts:
The [tex]\(x\)[/tex]-intercepts of a quadratic function [tex]\( f(x) = ax^2 + bx + c \)[/tex] are the values of [tex]\( x \)[/tex] where [tex]\( f(x) \)[/tex] equals zero. This means we need to solve the equation:
[tex]\[ 1.69x^2 + 11.7x + 20.25 = 0 \][/tex]
Using the quadratic formula, [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1.69 \)[/tex], [tex]\( b = 11.7 \)[/tex], and [tex]\( c = 20.25 \)[/tex], we can compute the roots.
After solving this equation, we find the [tex]\( x \)[/tex]-intercepts to be:
[tex]\[ x = -3.46, -3.46 \][/tex]
Since these are the roots, each value of [tex]\(-3.46\)[/tex] is repeated, indicating it is a double root.
### Finding the Vertex:
The vertex of a quadratic function in the form [tex]\( f(x) = ax^2 + bx + c \)[/tex] can be found using the formula for the [tex]\(x\)[/tex]-coordinate of the vertex, which is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Substituting the values [tex]\(a = 1.69\)[/tex] and [tex]\(b = 11.7\)[/tex]:
[tex]\[ x = -\frac{11.7}{2 \times 1.69} = -\frac{11.7}{3.38} = -3.46 \][/tex]
To find the [tex]\( y \)[/tex]-coordinate of the vertex, we substitute [tex]\( x = -3.46 \)[/tex] back into the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(-3.46) = 1.69(-3.46)^2 + 11.7(-3.46) + 20.25 \][/tex]
After calculating this, we find:
[tex]\[ f(-3.46) = 0.0 \][/tex]
So, the coordinates of the vertex are:
[tex]\[ (-3.46, 0.0) \][/tex]
### Summary:
The [tex]\( x \)[/tex]-intercepts are:
[tex]\[ -3.46, -3.46 \][/tex]
And the vertex of the quadratic function is:
[tex]\[ (-3.46, 0.0) \][/tex]
Therefore, the final answers are:
[tex]\(x\)[/tex]-intercept(s):
[tex]\[ -3.46 \][/tex]
vertex:
[tex]\[ (-3.46, 0.0) \][/tex]
