sketch the following graphs, under these conditions: ske share - Axis of symmetry - Uuning point - Interepts ( \( x \) and \( y \) ) - Maxmum or minimum values C.1. \( y=12 x^{2}+5 x-2 \) C.12. Does your graph intergect with \( x-9 x \) is How many times? (13. Determine tho discrimnant of \( )=12 x^{2}+5 x-2 \)

We will analyze the quadratic function   y = 12x² + 5x – 2 step‐by‐step, finding its important features and then investigating its intersections with the line y = x – 9. Finally, we will compute its discriminant. ────────────────────────────── 1. Finding the Axis of Symmetry For a quadratic in the form y = ax² + bx + c the axis of symmetry is given by   x = –b/(2a). Here a = 12 and b = 5, so   x = –5/(2·12) = –5/24. ────────────────────────────── 2. Finding the Turning Point (Vertex) The x-coordinate of the vertex is the same as the axis of symmetry, x = –5/24. To get the y-coordinate, substitute x = –5/24 into the function:   y = 12(–5/24)² + 5(–5/24) – 2. Step‑by‑step: a. Compute (–5/24)²:   (–5/24)² = 25/576. b. Multiply by 12:   12·(25/576) = 300/576 = 25/48   (since 300/576 reduces dividing numerator and denominator by 12). c. Compute the second term:   5(–5/24) = –25/24.   It is convenient to use a denominator of 48: –25/24 = –50/48. d. Express –2 as –96/48. Combine the terms:   y = (25/48) – (50/48) – (96/48)     = (25 – 50 – 96) / 48     = –121/48. Thus, the vertex is at   (–5/24, –121/48). Because a = 12 > 0, the parabola opens upward and the vertex represents a minimum point. ────────────────────────────── 3. Finding the Intercepts A. y‑intercept: Set x = 0.   y = 12·0² + 5·0 – 2 = –2. So, the y‑intercept is (0, –2). B. x‑intercepts: Solve 12x² + 5x – 2 = 0. Use the quadratic formula:   x = [ –b ± √(b² – 4ac) ] / (2a). First, calculate the discriminant for these intercepts:   Δ = 5² – 4·12·(–2) = 25 + 96 = 121. Then,   x = ( –5 ± √121 ) / (24)     = ( –5 ± 11 ) / 24. Thus, the two solutions are:   x = (–5 + 11)/24 = 6/24 = 1/4   x = (–5 – 11)/24 = –16/24 = –2/3. So the x‑intercepts are at (1/4, 0) and (–2/3, 0). ────────────────────────────── 4. Maximum or Minimum Value Since the coefficient a = 12 > 0, the parabola opens upward and so the vertex gives the minimum value of the function. The minimum value is y = –121/48 (approximately –2.52) which occurs at x = –5/24. ────────────────────────────── 5. Sketching the Graph When sketching the graph, include these features:  • The axis of symmetry: the vertical line x = –5/24.  • The turning point (vertex): (–5/24, –121/48).  • The x‑intercepts: (–2/3, 0) and (1/4, 0).  • The y‑intercept: (0, –2).  • Note that the graph opens upward (minimum at the vertex). A rough sketch would show a “U‑shaped” parabola with the vertex slightly left of the y‑axis, crossing the x‑axis at x ≈ –0.67 and x ≈ 0.25, and crossing the y‑axis at –2. ────────────────────────────── 6. Intersection with the Line y = x – 9 To see if the parabola and the line intersect, set their equations equal:   12x² + 5x – 2 = x – 9. Subtract (x – 9) from both sides:   12x² + 5x – 2 – x + 9 = 0    12x² + 4x + 7 = 0. Determine the discriminant for this quadratic:   Δ' = (4)² – 4(12)(7) = 16 – 336 = –320. Since Δ' is negative, there are no real solutions. Therefore, the parabola and the line y = x – 9 do not intersect at any real points. ────────────────────────────── 7. Discriminant of y = 12x² + 5x – 2 As computed earlier for the x‑intercepts:   Δ = 5² – 4·12·(–2) = 25 + 96 = 121. A positive discriminant (121 > 0) means the quadratic has two distinct real roots. ────────────────────────────── Summary of Answers: C.1 (Graph y = 12x² + 5x – 2):  • Axis of symmetry: x = –5/24.  • Vertex (turning point): (–5/24, –121/48), which is the minimum point.  • y‑intercept: (0, –2).  • x‑intercepts: (–2/3, 0) and (1/4, 0).  • Minimum value: y = –121/48. C.12 (Intersection with y = x – 9):  • Setting 12x² + 5x – 2 equal to x – 9 leads to 12x² + 4x + 7 = 0 with a discriminant of –320.  • Therefore, the graph of y = 12x² + 5x – 2 and the line y = x – 9 do not intersect (0 points of intersection). C.13 (Discriminant of y = 12x² + 5x – 2):  • The discriminant is 121. This completes the analysis and sketching of the quadratic graph according to the given conditions.