Answer:
A) x-intercepts are (0.6, 0) and (-1, 0)
B) vertex is (-0.2, -3.2)
C) see attached
Step-by-step explanation:
Part A
Given function: [tex]f(x)=5x^2+2x-3[/tex]
The x-intercepts are when f(x) = 0
[tex]\implies 5x^2+2x-3=0[/tex]
To factor, find two numbers that multiply to -15 and sum to 2: 5 and -3
Rewrite the middle term of the quadratic as the sum of these number:
[tex]\implies 5x^2+5x-3x-3=0[/tex]
Factorize the first two terms and the last two terms separately:
[tex]\implies 5x(x+1)-3(x+1)=0[/tex]
Factor out the common term [tex](x+1)[/tex]:
[tex]\implies (5x-3)(x+1)=0[/tex]
[tex]\implies (5x-3)=0\implies x=\dfrac35=0.6[/tex]
[tex]\implies (x+1)=0 \implies x=-1[/tex]
Therefore, the x-intercepts are (0.6, 0) and (-1, 0)
Part B
As the leading coefficient of the quadratic is positive, the parabola will open upwards. This means that the vertex will be a minimum point.
The x-coordinate of the vertex is the midpoint of the zeros.
[tex]\sf midpoint=\dfrac{-1+0.6}{2}=-0.2[/tex]
To find the y-coordinate of the vertex, substitute the found value of x into the given equation:
[tex]f(-0.2)=5(-0.2)^2+2(-0.2)-3=-3.2[/tex]
Therefore, the vertex is (-0.2, -3.2)
Part C
Plot the zeros and the vertex.
The axis of symmetry is the x value of the vertex, so ensure that the graph is symmetrical about x = -0.2.
