Answer:
(i) (0, -32)
(ii) (-4, 0) and (8, 0)
(iii) x = 2
(iv) (2, -36)
(v) (2, -36) minimum
Step-by-step explanation:
Given <u>quadratic equation</u>:
![y=x^2-4x-32](https://tex.z-dn.net/?f=y%3Dx%5E2-4x-32)
<u>Part (i)</u>
The <u>y-intercept</u> is the point at which the curve <u>crosses the y-axis</u>.
To find the y-intercept, substitute x = 0 into the given equation:
![\implies (0)^2-4(0)-32=-32](https://tex.z-dn.net/?f=%5Cimplies%20%280%29%5E2-4%280%29-32%3D-32)
Therefore, the y-intercept is at (0, -32)
<u>Part (ii)</u>
The <u>zeros</u> are the points at which the curve <u>crosses the x-axis</u>.
To find the zeros, substitute y = 0 into the given equation and factor:
![\implies x^2-4x-32=0](https://tex.z-dn.net/?f=%5Cimplies%20x%5E2-4x-32%3D0)
![\implies x^2-8x+4x-32=0](https://tex.z-dn.net/?f=%5Cimplies%20x%5E2-8x%2B4x-32%3D0)
![\implies x(x-8)+4(x-8)=0](https://tex.z-dn.net/?f=%5Cimplies%20x%28x-8%29%2B4%28x-8%29%3D0)
![\implies (x+4)(x-8)=0](https://tex.z-dn.net/?f=%5Cimplies%20%28x%2B4%29%28x-8%29%3D0)
Therefore:
![(x+4)=0 \implies x=-4](https://tex.z-dn.net/?f=%28x%2B4%29%3D0%20%5Cimplies%20x%3D-4)
![(x-8)=0 \implies x=8](https://tex.z-dn.net/?f=%28x-8%29%3D0%20%5Cimplies%20x%3D8)
So the zeros are (-4, 0) and (8, 0)
<u>Part (iii)</u>
The <u>axis of symmetry</u> is a vertical straight line that divides the curve into two symmetrical parts. The axis of symmetry is the x-value of the mid-point of the zeros.
![\sf \implies midpoint=\dfrac{8+(-4)}{2}=2](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20midpoint%3D%5Cdfrac%7B8%2B%28-4%29%7D%7B2%7D%3D2)
Therefore, the axis of symmetry is: x = 2
<u>Part (iv)</u>
The <u>vertex</u> is the <u>turning point</u> of the parabola.
If the leading coefficient is <u>positive</u>, the parabola opens <u>upwards</u> and the vertex is the <u>minimum</u> point.
If the leading coefficient is <u>negative</u>, the parabola opens <u>downwards</u> and the vertex is the <u>maximum</u> point.
The axis of symmetry is the x-value of the vertex.
To find the y-value, substitute x = 2 into the equation:
![\implies (2)^2-4(2)-32=-36](https://tex.z-dn.net/?f=%5Cimplies%20%282%29%5E2-4%282%29-32%3D-36)
Therefore, the vertex is (2, -36)
<u>Part (v)</u>
The <u>optimal value</u> is also known as the <u>vertex</u>.
Therefore, the optimal value is (2, -36).
As the leading coefficient of the given quadratic equation is <u>positive</u>, the parabola opens <u>upwards</u> and so the optimal value is a minimum.