Answer :
To find the [tex]\( x \)[/tex]-intercepts of the function [tex]\( f(x) = x^2 + 5x - 36 \)[/tex], we need to solve the equation [tex]\( f(x) = 0 \)[/tex] or [tex]\( x^2 + 5x - 36 = 0 \)[/tex].
1. Identify coefficients:
The given quadratic equation is [tex]\( x^2 + 5x - 36 = 0 \)[/tex]. Here, the coefficients are:
[tex]\[ a = 1, \quad b = 5, \quad c = -36 \][/tex]
2. Calculate the discriminant:
The discriminant of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = 5^2 - 4(1)(-36) = 25 + 144 = 169 \][/tex]
3. Find the roots using the quadratic formula:
The quadratic formula to find the roots of [tex]\( ax^2 + bx + c = 0 \)[/tex] is:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( \Delta \)[/tex]:
[tex]\[ x = \frac{-5 \pm \sqrt{169}}{2 \cdot 1} = \frac{-5 \pm 13}{2} \][/tex]
4. Calculate the two potential solutions:
[tex]\[ x_1 = \frac{-5 + 13}{2} = \frac{8}{2} = 4 \][/tex]
[tex]\[ x_2 = \frac{-5 - 13}{2} = \frac{-18}{2} = -9 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercepts of the graph [tex]\( f(x) = x^2 + 5x - 36 \)[/tex] are the points where the graph intersects the [tex]\( x \)[/tex]-axis, which are [tex]\( (4, 0) \)[/tex] and [tex]\( (-9, 0) \)[/tex].
Therefore, the correct pair is:
[tex]\[ (4, 0) \text{ and } (-9, 0) \][/tex]
1. Identify coefficients:
The given quadratic equation is [tex]\( x^2 + 5x - 36 = 0 \)[/tex]. Here, the coefficients are:
[tex]\[ a = 1, \quad b = 5, \quad c = -36 \][/tex]
2. Calculate the discriminant:
The discriminant of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = 5^2 - 4(1)(-36) = 25 + 144 = 169 \][/tex]
3. Find the roots using the quadratic formula:
The quadratic formula to find the roots of [tex]\( ax^2 + bx + c = 0 \)[/tex] is:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( \Delta \)[/tex]:
[tex]\[ x = \frac{-5 \pm \sqrt{169}}{2 \cdot 1} = \frac{-5 \pm 13}{2} \][/tex]
4. Calculate the two potential solutions:
[tex]\[ x_1 = \frac{-5 + 13}{2} = \frac{8}{2} = 4 \][/tex]
[tex]\[ x_2 = \frac{-5 - 13}{2} = \frac{-18}{2} = -9 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercepts of the graph [tex]\( f(x) = x^2 + 5x - 36 \)[/tex] are the points where the graph intersects the [tex]\( x \)[/tex]-axis, which are [tex]\( (4, 0) \)[/tex] and [tex]\( (-9, 0) \)[/tex].
Therefore, the correct pair is:
[tex]\[ (4, 0) \text{ and } (-9, 0) \][/tex]
