The equation of the parabola is given to be:
[tex]y=x^2+6x-17[/tex]X-INTERCEPTS
To find the x-intercepts, we can substitute y = 0 in the equation. This gives us:
[tex]x^2+6x-17=0[/tex]This is a quadratic equation. To solve it, we can use the Quadratic Formula given as:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]where a and b are the coefficients of the variables with power 2 and 1 respectively and c is the constant term.
We can use the following parameters to solve the question:
[tex]\begin{gathered} a=1 \\ b=6 \\ c=-17 \end{gathered}[/tex]Inputting into the formula, we have:
[tex]\begin{gathered} x=\frac{-6\pm\sqrt[]{6^2-(4\times1\times\lbrack-17\rbrack)}}{2\times1}=\frac{-6\pm\sqrt[]{36+68}}{2} \\ x=\frac{-6\pm\sqrt[]{104}}{2}=\frac{-6\pm10.20}{2} \end{gathered}[/tex]Therefore, the values for x can be:
[tex]x=\frac{-6+10.20}{2}=2.10[/tex]or
[tex]x=\frac{-6-10.20}{2}=-8.10[/tex]Therefore, the x-intercepts are (-8.10, 0) and (2.10, 0).
Y-INTERCEPT
The y-intercept of the parabola can be gotten by substituting x = 0 into the equation as shown below:
[tex]\begin{gathered} y=(0)^2+6(0)-17 \\ y=-17 \end{gathered}[/tex]The y-intercept of the parabola is (0, -
